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Indefinite Integration

1. An antiderivative of a function y = f(x) defined on some interval (a, b) is called any

function F(x) whose derivative at any point of this interval is equal to f(x):

F'(x) = f(x).

If F(x) is an antiderivative of f(x), then the function of the form F(x) + C, where C is

an arbitrary constant, is also an antiderivative of f(x).

2. The indefinite integral of a function f(x) is the collection of all antiderivatives for

this function:

3. The derivative of the indefinite integral is equal to the integrand:

4. The indefinite integral of the sum of two functions is equal to the sum of the

integrals:

5. The indefinite integral of the difference of two functions is equal to the difference

of the integrals:

6. A constant factor can be moved across the integral sign:

7.

8.

9.

10.

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11. Integration by substitution

12. Integration by parts

where u(x), v(x) are differentiable functions.

Definition of the Antiderivative and Indefinite Integral

The function F(x) is called an antiderivative of f(x), if

The family of all antiderivatives of a function f(x) is called the indefinite integral of the function f(x)

and is denoted by

Thus, if F is a particular antiderivative, we may write

where C is an arbitrary constant.

Properties of the Indefinite Integral

In formulas given below f and g are functions of the variable x, F is an antiderivative of f, and

a, k, C are constants.

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Table of Integrals

It's supposed below that a, p (p ≠ 1), C are real constants, b is the base of the exponential function

(b ≠ 1, b > 0).

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Q.1. Calculate

. Solution.

Q.2. Calculate the integral

. Solution.

Transforming the integrand and using the formula for integral of the power function, we have

Q.3. Calculate

. Solution.

We use the table integral

.

Then

Q.4. Find the integral

. Solution.

Using the table integral

,

we obtain

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. Solution.

Since , the integral is


. Solution.

Using the double angle formula sin 2x = 2 sin x cos x and the identity sin2x + cos

2x = 1, we can write:

Q.7. Evaluate the integral

. Solution.

Decompose the integrand into partial functions:

Equate coefficients:

Hence,

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Then

The integral is equal to

Q.8. Evaluate

. Solution.

First we divide the numerator by the denominator, obtaining

Then


. Solution.

We can write:


. Solution.


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Hence,

Then


Q.11.Evaluate

. Solution.

Decompose the integrand into the sum of two fractions:


Hence

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The integrand can be written as

The initial integral becomes


. Solution.

We can factor the denominator in the integrand:



Hence,

Then

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Now we can calculate the initial integral:


. Solution.

Rewrite the denominator in the integrand as follows:

The factors in the denominator are irreducible quadratic factors since they have no real roots. Then


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This yields

Hence,

Integrating term by term, we obtain the answer:

We can simplify this answer. Let

Then

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Hence, . The complete answer is


. Solution.

We can factor the denominator in the integrand:



Hence,

Thus, the integrand becomes

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So, the complete answer is

Q.15.Calculate the integral

. Solution.

Decompose the integrand into partial functions, taking into account that the denominator has a

third degree root:


We get the following system of equations:

Hence,

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The initial integral is equal to


. Solution.

Since is reducible, we complete the square in the denominator:

Now, we can compute the integral using the reduction formula

Then


. Solution.

We make the substitution:

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Then


. Solution.

We make the following substitution:

Then the integral (we denote it by I ) becomes

Divide the fraction:

As a result, we have


. Solution.

We can write the integral as

Since the least common multiple (LCM) of the denominators of the fractional powers is equal

to n = LCM(1,3) = 3, we make the substitution:

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Thus

Make the new substitution:

The final answer is


. Solution.


Make the substitution:

The integral becomes

Since the degree of the numerator is greater than the degree of the denominator,

we divide the numerator by the denominator:


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. Solution.


As can be seen, the least common multiple (LCM) of the denominators of the fractional powers

is equal to n = LCM(3,4) = 12, we make the substitution:

This yields

The degree of the numerator is greater than the degree of the denominator, therefore,

we divide the fraction:

After simple transformations we obtain the final answer:

Q.22. Evaluate

. Solution.


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This yields

Q.23.Calculate the integral

. Solution.

We make the substitution



. Solution.

We use the universal trigonometric substitution:

Since

,

we have


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. Solution.

Make the universal trigonometric substitution:

Then the integral becomes


. Solution.

As in the previous examples, we will use the universal trigonometric substitution:

Since we can write:

Q.27. Evaluate

. Solution.

We can write the integral in form:

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Use the universal trigonometric substitution:

This leads to the following result:


. Solution.

Since , we can write

Hence,

so the integral can be transformed in the following way:

Make the substitution

.

Then use the identity

We get the final answer:

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. Solution.

We solve this integral by making the trigonometric substitution

Taking into account that

we find the integral:


. Solution.


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As a result, the integral becomes

Using the method of partial fractions, we can write the integrand as

Calculate the coefficients A, B and C:

Hence,

So the integral is

Integration of Some Special Classes of Trigonometric Functions

In this section we consider 8 classes of integrals with trigonometric functions. Special transformations

and subtitutions used for each of these classes allow us to obtain exact solutions for these integrals.

1. Integrals of the form

To find integrals of this type, use the following trigonometric identities:

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It's assumed here and below that m and n are positive integers. To find an integral of this form,

use the following substitutions:

a. If the power n of the cosine is odd (the power m of the sine can be arbitrary),

then the substitution is used.

b. If the power m of the sine is odd, then the substitution is used.

c. If both powers m and n are even, then first use the double angle formulas

to reduce the power of the sine or cosine in the integrand.

Then, if necessary, apply the rules a) or b).


The power of the integrand can be reduced by using the trigonometric identity

and the reduction formula


We can reduce the power of the integrand using the trigonometric identity

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and the reduction formula


This type of integrals can be simplified with help of the reduction formula:


Similarly to the previous examples, this type of integrals can be simplified by the formula


a. If the power of the secant n is even, then using the identity

the secant function is expressed asthe tangent function. The factor is separated and used for

transformation of the differential. As a result,the entire integral (including differential) is expressed

as the function of tan x.

b. If both the powers n and m are odd, then the factor sec x tan x, which is necessary to transform the

differential, is separated. Then the entire integral is expressed through sec x.

c. If the power of the secant n is odd, and the power of the tangent m is even, then the tangent is

d. expressed as the secant using the identity . Then the integrals of the secant

are calculated.


a. If the power of the cosecant n is even, then using the identity the

cosecant function is expressed as the cotangent function. The factor is separated and

used for transformation of the differential. As a result, the integrand and differential are expressed

through cot x.

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b. If both the powers n and m are odd, then the factor csc x cot x, which is necessary to transform the

differential, is separated. Then the integral is expressed through csc x.

c. If the power of the cosecant n is odd, and the power of the cotangent m is even, then the cotangent

is expressed as the cosecant using the identity .

Then the integrals of the cosecant are calculated.


. Solution.

Let u = cos x, du = − sin xdx. Then


. Solution.

Making the substitution u = sin x, du = cos xdx and using the identity , we obtain


. Solution.

Using identities and , we can write:

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Calculate the integrals in the latter expression.

To find the integral , we make the substitution u = sin 2x, du = 2cos 2xdx. Then

Hence, the initial integral is


. Solution.

We can write:

Transform the integrand using the identities

We get


. Solution.

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Making the substitution u = cos x, du = − sin xdx and expressing the sine through cosine with help of the

formula , we obtain


. Solution.

Transform the integrand by the formula

Hence,



. Solution.

We use the identity

to transform the integral. This yields

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. Solution.

Using the identity , we have


. Solution.

We use the reduction formula

Hence,

The integral is a table integral which is equal to .

(It can be easily found usingthe universal trigonometric substitution .)

As a result, the integral becomes


. Solution.

We use the reduction formula

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Hence,

Q.41.Compute

. Solution.

Q.42. Compute

. Solution.

Use the identity

. Then

Since

and

is a table integral equal to

,

we obtain the following complete answer:

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. Solution.


Then


. Solution.

To find the integral, we make the substitution:

.

Using the identity

,

we have

Express sin t in terms of x:

Hence, the integral is

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. Solution.


.

Hence,


Returning back to the variable x :

we obtain the complete answer:


. Solution.

We make the trigonometric substitution:

x = a sec t, dx = a tan t sec t dt.

Calculate the integral using the identity

:

For t, we have the following expression:

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Hence, returning back to the variable x, we obtain:


. Solution.


Make the trigonometric substitution:

Now we can calculate the integral:


. Solution.

First we complete the square in the integrand.

Then, making the substitution

and using the identity

,

we can write:

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. Solution.


Then

Express and in terms of x:

Hence,

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. Solution.

Using the trigonometric substitution

we get

Now, we make the substitution . The integral becomes

Use the method of partial fractions to transform the integrand:

Equate the coefficients:

Then

Hence, we can write the integrand as

So, the initial integral is equal to

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Returning back to the variable x, we have

Hence, the final answer is

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The Definite Integral

Let f (x) be a continuous function on the closed interval [a, b].

The definite integral of f (x) from a to b is defined to be the limit

where

THE LIMIT DEFINITION OF A DEFINITE INTEGRAL

The following problems involve the limit definition of the definite integral of a continuous function

of one variable on a closed, bounded interval. Begin with a continuous function

on the interval

Let

...

be an arbitrary (randomly selected) partition of the interval , which divides the interval into subintervals (subdivisions). Let

...

be the sampling numbers (or sampling points) selected from the subintervals. That is,

is in ,

is in ,

is in , ... ,

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is in ,

is in , and

is in .

Define the mesh of the partition to be the length of the largest subinterval. That is, let

for and define

.

The definite integral of on the interval is most generally defined to be

.

For convenience of computation, a special case of the above definition uses subintervals

of equal length and sampling points chosen to be the right-hand endpoints of the subintervals.

Thus, each subinterval has length

equation

for and the right-hand endpoint formula is

equation

for . The definite integral of on the interval can now be alternatively

defined by

.

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We will need the following well-known summation rules.

1. (n times) , where is a constant

2.

3.

4.

5. , where is a constant

6.

PROBLEM 1 : Use the limit definition of definite integral to evaluate

.


.


.


.


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.


.


.

SOLUTIONS TO THE LIMIT DEFINITION OF A DEFINITE INTEGRAL

SOLUTION 1 : Divide the interval into equal parts each of length

for . Choose the sampling points to be the right-hand endpoints of the subintervals and given by

for . The function is

.

Then the definite integral is

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(Since is the variable of the summation, the expression is a constant.

Use summation rule 1 from the beginning of this section.)

.


for . Choose the sampling points to be the right-hand endpoints of the

subintervals and given by


.


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(Use summation rule 6 from the beginning of this section.)

(Use summation rules 5 and 1 from the beginning of this section.)


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.





.


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.




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.





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.





.


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.

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.




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.

Properties of the Definite Integral

We assume below that f (x) and g (x) are continuous functions on the closed interval [a, b].

1.

2. where k is a constant;

3.

4.

5. If for all , then .

6.

7.

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8. If in the interval [a, b], then

The Fundamental Theorem of Calculus

Let f (x) be a function, which is continuous on the closed interval [a, b].

If F (x) is any antiderivative of f (x) on [a, b],then

The Area under a Curve

The area under the graph of the function f (x) between the vertical lines x = a, x = b (Figure 1)

is given by the formula

Fig.1

Fig.2

Let F (x) and G (x) be indefinite integrals of functions f (x) and g (x), respectively.

If f (x) ≥ g (x) on the closed interval [a, b], then the area between the curves y = f (x), y = g (x) and

the lines x = a, x = b (Figure 2) is given by

The Method of Substitution for Definite Integrals

The definite integral

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of the variable x can be changed into an integral with respect to t by making the substitution x = g (t):

The new limits of integration for the variable t are given by the formulas

where g -1

is the inverse function to g, i.e. t = g -1

(x).

Integration by Parts for Definite Integrals

In this case the formula for integration by parts looks as follows:

where means the difference between the product of functions uv at x = b and x = a.


. Solution.

Using the fundamental theorem of calculus, we have


. Solution.

Q.3.Evaluate the integral

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. Solution.

First we make the substitution:

Determine the new limits of integration. When x = 0, then t = −1.

When x = 1, then we have t = 2. So, the integral with the new variable t can be easily calculated:


. Solution.

We can write

Apply integration by parts:

.

In this case, let

Hence, the integral is

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Q.5. Find the area bounded by the curves and . Solution.

First we find the points of intersection (see Figure 3):

Fig.3

As can be seen, the curves intercept at the points (0,0) and (1,1). Hence, the area is given by

Q.6. Find the area bounded by the curves and . Solution.

First we find the points of intersection of the curves (Figure 4):

The upper boundary of the region is the parabola

and the lower boundary is the straight line

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The area is given by

Q.8. Find the area inside the ellipse

. Solution.

By symmetry (see Figure 6), the area of the ellipse is twice the area above the x-axis.

The latter is given by

To calculate the last integral, we use the trigonometric substitution x = asin t, dx = acos tdt.

Refine the limits of integration. When x = − a, then sin t = −1, and .

When x = a, then sin t = 1, . Thus we get

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Hence, the total area of the ellipse is πab.

Try to Your Self

Q. 1. Find the area lying above the x-axis and included between the circle x2+y2=8x and the parabola y2 =4x.

Q. 2. Find the area lying above the x-axis and included between the circle x2 +y2=16a2 and the parabola y2 =6ax.

Q. 3. Find the area of the smaller region bounded by the ellipse and the

line

Q. 4. Find the area of the region included between x2 =4y , y = 2 , y = 4 and the y-axis in the first quadrant.

Q. 5. Find the area between the parabolas 4ay = x2 and y2 = 4ax.

Q. 6. Find the area bounded by the curve y2 = 4ax and the line y = 2a and y-axis.

Q. 7. Find the area bounded by the parabola y2 = 8x and its latus rectum

Q. 8. Find the area of the circle x2 + y2 = 16, which is exterior to the parabola y2 = 6x.

Q. 9. Sketch the region common to the circle x2 + y2 = 8 and the parabola x2 = 4y. Also find the area of the common region using integration.

Q. 10. Draw the rough sketch of the region and find the area enclosed by the region using method of integration.

Q. 11. Using integration, find the area of the triangle ABC whose vertices are A(2,3), B(2,8) and C(6,5).

Q. 12. Using integration, find the area of the triangle ABC whose vertices are A(2,5), B(4,7) and C(6,2).

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Q. 13. Using integration, find the area of the triangle ABC whose vertices are A(-1,1), B(0.5) and C(3,2).

Q. 14. Compute the area bounded by the lines x+2y = 2, y-x=1 and 2x+y = 7.

Q. 15. Compute the area bounded by the lines y = 4x+5, y = 5-x, and 4y = x+5.

Q. 16. Compute the area bounded by the lines 2x+y = 4, 3x-2y = 6, and x-3y+5=0.

Q. 17. Using integration, find the area of the region bounded by x-7y+19=0, and y =çxú.

Q. 18. Using integration, find the area of the region bounded by the line

i. 2y= -x+8, x-axis and the lines x = 2 and x = 4. ii. y -1 = x, x-axis and the lines x = -2 and x = 3

iii. y = , line y = x and the positive x- axis.

Q. 19. Find the area of the region enclosed between the two circles x2 + y2 = 1 and (x-4)2 + y2 =16.

Q. 20. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = 4x

and the circle x2 + y2 = 32

Q. 21. Find the area of the smaller part of the circle x2 + y2 =a2 cut off by

the line .

Q. 22. Find the area of the region

and evaluate Q. 23. Sketch the graph of the curve y =

Q. 24. Sketch the graph of the curve and find the area bounded by y = , x=-2, x=3,

y=0.

Q. 25. Find the area bounded by the line y = sin2x and y = cos2x between x = 0 and x=p/4

Q. 26. Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded

by x = 0, x = 4, y = 4 and y = 0 into three equal parts.

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Differential Equation

Separable Equations

A first order differential equation y' = f(x,y) is called a separable equation if the function f(x,y) can

be factored into the product of two functions of x and y:

where p(x) and h(y) are continuous functions.

Considering the derivative y' as the ratio of two differentials we move dx to the right side

and divide the equation by h(y):

Of course, we need to make sure that h(y) ≠ 0. If there's a number x0 such that h(x0) = 0,

then this number will also be a solution of the differential equation. Division by h(y) causes

loss of this solution.

Denoting , we write the equation in the form

We have separated the variables so now we can integrate this equation:

where C is an integration constant.

Calculating the integrals, we get the expression

representing the general solution of the separable differential equation.

Q.1. Solve the differential equation

. Solution.

In the given case p(x) = 1 and h(y) = y(y +2). Divide the equation by h(y) and move dx to the right side:

One can notice that by dividing we can lose the solutions y = 0 and y = −2 when h(y) becomes zero.

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In fact, let's see that y = 0 is a solution of the differential equation. Obviously,

Substituting this into the equation gives 0 = 0. Hence, y = 0 is one of the solutions. Similarly,

we can check thaty = −2 is also a solution.

Returning to the differential equation, we integrate it:

We can calculate the left integral by using the fractional decomposition of the integrand:

Thus, we get the following decomposition of the rational integrand:

Hence,

We can rename the constant: 2C = C1. Thus, the final solution of the equation is written in the form

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Here the general solution is expressed in implicit form. In the given case we can transform

the expression to obtain the answer as an explicit function y = f(x,C1), where C1 is a constant.

However, it's possible to do not for all differential equations.


. Solution.

We can rewrite this equation in the following way:

Divide both sides by (x2 + 4)y to get

Obviously, that x2 + 4 ≠ 0 for all real x. Check if y = 0 is a solution of the equation.

Substituting y = 0 and dy = 0 into the differential equation, we see that the function y = 0

is one of the solutions of the equation.

Now we can integrate it:

Notice that dx2 = d(x

2 + 4). Hence,

We can represent the constant C as lnC1, where C1 > 0. Then

Thus, the given differential equation has the following solutions:

This answer can be simplified. Indeed, if using an arbitrary constant C, which takes values

from −∞ to +∞, the solution can be written in the form:

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When C = 0, it becomes y = 0.

Q.3. Find all solutions of the differential equation y' = −xe y.

Solution.

Transform this equation in following way:

Obviously, that division by e y does not lead to the loss of solutions, since e

y > 0. After integrating we have

This answer can be expressed in the explicit form:

We assume in the latter expression that the constant C > 0 in order to satisfy the domain of

the logarithmic function.

Q.4. Find a particular solution of the differential equation

under condition y(1) = −1. Solution.

Divide both sides of the equation by x:

We suppose that x ≠ 0, because the domain of the given equation is x > 0.

Integrating this equation yields:

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The integral in the right side is calculated as follows:

Hence, the general solution in the implicit form is given by

where C1 = 2C is an integration constant. Find the values of C1 to satisfy the initial condition y(1) = −1:

Thus, the particular solution satisfying the initial condition is written in the following way:

Q.5. Solve the differential equation y' cot2 x + tan

2 y = 0.

Solution.

We write this equation as follows:

Divide both sides of the equation by tan y cot2 x:

Check for possible missed solutions when dividing. There might be the following two roots:

Substituting this into the initial equations, we see that is a solution.

The second possible solution is given by

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Here we get the answer

which does not satisfy the initial differential equation.

Now we can integrate the given equation and find its general solution:

The final answer is given by

Q.6. Find a particular solution of the equation

satisfying the initial condition y(0) = 0. Solution.

Write this equation in the following way:

Divide both sides by 1 + ex:

Since 1 + ex > 0, then we did not miss solutions of the original equation. Integrating this equation yields:

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Determine the constant C from the initial condition y(0) = 0.

Hence, the final answer is given by

Q.7. Solve the equation

. Solution.

The product xy in each side does not allow separating the variables. Therefore, we make the replacement:

The relationship for differential is given by

Substituting this into the equation, we can write:

By multiplying both sides by x and then canceling the corresponding fractions in the left and right side,

we get

Take into account that x = 0 is a solution of the equation, which can be checked by direct substitution.

Simplify the latter expression:

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Now the variables x and t are separated:

After integrating we have

By making the reverse substitution t = xy, we find the general solution of the equation:

The complete answer is written in the form:

Homogeneous Equations

Definition of Homogeneous Differential Equation

A first order differential equation

is called homogeneous equation, if the right side satisfies the condition

for all t. In other words, the right side is a homogeneous function (with respect to the variables x and y)

of the zero order:

A homogeneous differental equation can be also written in the form

or alternatively, in the differential form:

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where P(x,y) and Q(x,y) are homogeneous functions of the same degree.

Definition of Homogeneous Function

A function P(x,y) is called a homogeneous function of the degree n if the following relationship is

valid for all t > 0:

Solving Homogeneous Differential Equations

A homogeneous equation can be solved by substitution y = ux, which leads to a separable differential

equation.


.

Solution.

It is easy to see that the polynomials P(x,y) and Q(x,y), respectively, at dx and dy are homogeneous

functions of the first order. Therefore, the original differential equation is also homogeneous.

Suppose that y = ux, where u is a new function depending on x. Then

Substituting this into the differential equation, we obtain

Hence,

Dividing both sides by x yields:

When dividing by x, we could lose the solution x = 0. The direct substitution shows that x = 0 is

indeed a solution of the given differential equation.

Integrate the latter expression to obtain:

where C is a constant of integration.

Returning to the old variable y, we can write:

Thus, the equation has two solutions:

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.

Solution.

We notice that the root x = 0 does not belong to the domain of the differential equation.

Rewrite the equation in the form:

As seen, this equation is homogeneous.

Make the substitution y = ux. Hence,

Substituting this expression into the equation gives:

Divide by x ≠ 0 to get:

We obtain the separable equation:

The next step is to integrate the left and the right side of the equation:

Hence,

Here the constant C can be written as ln C1 (C1 > 0). Then

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Thus, we have got two solutions:

If C1 = 0, the answer is y = xe and we can make sure that it is also a solution to the equation.

Indeed, substituting

into the differential equation, we obtain:

Then all the solutions can be represented by one formula:

where C is an arbitrary real number.


.

Solution.

Here we deal with a homogeneous equation. In fact, we can rewrite it in the form:

Make the substitution y = ux. Then y' = u'x + u. Substituting y and y' into the original equation, we have

Divide both sides by ux2. We notice that x =0 is not the solution of the equation. However, one can

check that u =0or y =0 is one of the solutions of the differential equation.


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Integrating, we find the general solution:

Taking into account that , we can write the last expression in the form

The given expression can be represented as an explicit inverse function x(y):

Since C is an arbitrary real number, we can replace the "minus" sign before the constant to the "plus" sign.

Then

Thus, the given differential equation has the solutions:

Example 4

Solve the differential equation .

Solution.

As it follows from the right side of the equation, x ≠ 0 and y ≠ 0. We can make the substitution

y = ux, y' = u'x + u. This yields:

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Integrating this equation, we obtain:

Let the constant 2C be denoted by just C. Hence,

Thus, the general solution is written in the form

Q.5. Find the general solution of the differential equation

.

Solution.

It is easy to see that the given equation is homogeneous. Therefore, we can use the substitution

y = ux, y' = u'x + u. As a result, the equation is converted into the separable differential equation:

Divide both sides by x3. (Notice that x = 0 is not the solution).

Now we can integrate the last equation:

Since u = y/x, the solution can be written in the form:

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It follows from here that

We can denote eC = C1, (C1 > 0). Then the solution in the implicit form is given by the equation

where the constant C1 > 0.

Linear Differential Equations of First Order

Definition of Linear Equation of First Order

A differental equation of type

where a(x) and f(x) are continuous functions of x, is called a linear nonhomogeneous differential

equation of first order. We consider two methods of solving linear differential equations of first order:

Using an integrating factor;

Using an Integrating Factor

If a linear differential equation is written in the standard form:

the integrating factor is defined by the formula

Multiplying the left side of the equation by the integrating factor u(x) converts the left side into the

derivative of the product y(x)u(x).

The general solution of the differential equation is expressed as follows:


Method of Variation of a Constant

This method is similar to the previous approach. First it's necessary to find the general solution of

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the homogeneous equation:

The general solution of the homogeneous equation contains a constant of integration C. We replace

the constant C with a certain (still unknown) function C(x). By substituting this solution into the

nonhomogeneous differential equation, we can determine the function C(x).

The described algorithm is called the method of variation of a constant. Of course, both methods lead

to the same solution.

Initial Value Problem

If besides the differential equation, there is also an initial condition in the form of y(x0) = y0, such

a problem is called the initial value problem (IVP) or Cauchy problem.

A particular solution for an IVP does not contain the constant C, which is defined by substitution of

the general solution into the initial condition y(x0) = y0.

Q.1. Solve the equation y' − y − xex = 0.

Solution.

Rewrite this equation in standard form:

We will solve this equation using the integrating factor

Then the general solution of the linear equation is given by


.

Solution.

We will solve this problem by using the method of variation of a constant. First we find the general

solution of the homogeneous equation:

which is solved by separating the variables:

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where C is a positive real number.

Now we replace C with a certain (still unknown) function C(x) and will find a solution of the original

nonhomogeneous equation in the form:

Then the derivative is given by

Substituting this into the equation gives:

Upon integration, we find the function C(x):

where C1 is an arbitrary real number.

Thus, the general solution of the given equation is written in the form

Q.3. Solve the equation y' − 2y = x.

Solution.

First we solve this problem using an integrating factor. The given equation is already written in

the standard form. Therefore

Then the integrating factor is

The general solution of the original differential equation has the form:

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We calculate the last integral with help of integration by parts.

Then

Q.4. Solve the differential equation x2y' + xy + 2 = 0.

Solution.

We solve this problem using the method of variation of a constant. For convenience, we write

this equation in the standard form:

Divide both sides by x2. Obviously, that x = 0 is not the solution of the equation.

Consider the homogeneous equation:

After easy transformations we find the answer y = C/x, where C is any real number.

The last expression includes the case y = 0, which is also a solution of the homogeneous equation.

Now we replace the constant C with the function C(x) and substitute the solution y = C(x)/x into

the initial nonhomogeneous differential equation. Since

we obtain

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Thus, the general solution of the initial equation is given by

Q.5. Solve the initial value problem:

Solution.

First of all we calculate the integrating factor, which is written as

Here

Hence, the integrating factor is given by

We can take the function u(x) = cos x as the integrating factor. One can make sure that the left side

of the equation is the derivative of the product y(x)u(x):

Then the general solution of the given equation is written in the form:

Next we determine the value of C, which satisfies the initial condition y(0) = 1:

It follows from here that C = 4/3.

Hence, the solution for the initial value problem is given by

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Q.6. Solve the differential equation (IVP)

with the initial condition y(1) = 2.

Solution.

Determine the integrating factor:

We can take the function u(x) = x3 as the integrating factor. One can check that the left side of the equation is the

derivative of the product y(x)u(x):

The general solution of the differential equation is written as

Now we can find the constant C using the initial condition y(1) = 2. Substituting the general solution into this

condition gives:

Thus, the solution of the IVP is given by

Q.7. Find the general solution of the differential equation y = (2y4 + 2x)y'.

Solution.

It can be seen that this equation is not linear with respect to the function y(x). However, we can try to

find the solution for the inverse function x(y). We write the given equation through differentials and

make some transformations:

Now we see that we have a linear differential equation with respect to the function x(y).

We can solve it with help of the integrating factor:

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Then the general solution as the inverse function x(y) is given by




where a(x) and f(x) are continuous functions of x, is called a linear nonhomogeneous differential equation of first

order. We consider two methods of solving linear differential equations of first order:


Method of variation of a constant.




Multiplying the left side of the equation by the integrating factor u(x) converts the left side into the derivative of the

product y(x)u(x).




This method is similar to the previous approach. First it's necessary to find the general solution of the homogeneous

equation:

The general solution of the homogeneous equation contains a constant of integration C. We replace the

constant Cwith a certain (still unknown) function C(x). By substituting this solution into the nonhomogeneous

differential equation, we can determine the function C(x).

The described algorithm is called the method of variation of a constant. Of course, both methods lead to the same

solution.


If besides the differential equation, there is also an initial condition in the form of y(x0) = y0, such a problem is

called the initial value problem (IVP) or Cauchy problem.

A particular solution for an IVP does not contain the constant C, which is defined by substitution of the general

solution into the initial condition y(x0) = y0.

Example 1

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Solve the equation y' − y − xex = 0.

Solution.




Example 2


Solution.

We will solve this problem by using the method of variation of a constant. First we find the general solution of the

homogeneous equation:







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Example 3

Solve the equation y' − 2y = x.

Solution.

A. First we solve this problem using an integrating factor. The given equation is already written in the standard form.

Therefore




Then

B. Now we construct the solution by the method of variation of a constant. Consider the corresponding homogeneous

equation:

and find its general solution.

where C again denotes any real number. Notice that at C = 0, we get y = 0 that is also a solution of the homogeneous

equation.

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Next we suppose that C is a function of x and substitute the solution y = C(x)e2x into the initial nonhomogeneous

equation. We can write

Hence,

This integral was already found above in section A, so we obtain

As a result, the general solution of the nonhomogeneous differential equation is given by

As you can see, both methods give the same answer :).

Example 4

Solve the differential equation x2y' + xy + 2 = 0.

Solution.

We solve this problem using the method of variation of a constant. For convenience, we write this equation in the

standard form:



After easy transformations we find the answer y = C/x, where C is any real number. The last expression includes the

case y = 0, which is also a solution of the homogeneous equation.

Now we replace the constant C with the function C(x) and substitute the solution y = C(x)/x into the initial

nonhomogeneous differential equation. Since

we obtain

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Example 5

Solve the initial value problem:

Solution.


Here


We can take the function u(x) = cos x as the integrating factor. One can make sure that the left side of the equation is

the derivative of the product y(x)u(x):





Example 6

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Solve the differential equation (IVP) with the initial condition y(1) = 2.

Solution.


We can take the function u(x) = x3 as the integrating factor. One can check that the left side of the equation is the

derivative of the product y(x)u(x):


Now we can find the constant C using the initial condition y(1) = 2. Substituting the general solution into this

condition gives:


Example 7

Find the general solution of the differential equation y = (2y4 + 2x)y'.

Solution.

It can be seen that this equation is not linear with respect to the function y(x). However, we can try to find the

solution for the inverse function x(y). We write the given equation through differentials and make some

transformations:

Now we see that we have a linear differential equation with respect to the function x(y). We can solve it with help of

the integrating factor:


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where a(x) and f(x) are continuous functions of x, is called a linear nonhomogeneous differential equation of first

order. We consider two methods of solving linear differential equations of first order:


Method of variation of a constant.




Multiplying the left side of the equation by the integrating factor u(x) converts the left side into the derivative of the

product y(x)u(x).




This method is similar to the previous approach. First it's necessary to find the general solution of the homogeneous

equation:

The general solution of the homogeneous equation contains a constant of integration C. We replace the

constant Cwith a certain (still unknown) function C(x). By substituting this solution into the nonhomogeneous

differential equation, we can determine the function C(x).

The described algorithm is called the method of variation of a constant. Of course, both methods lead to the same

solution.


If besides the differential equation, there is also an initial condition in the form of y(x0) = y0, such a problem is

called the initial value problem (IVP) or Cauchy problem.

A particular solution for an IVP does not contain the constant C, which is defined by substitution of the general

solution into the initial condition y(x0) = y0.

Example 1

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Solve the equation y' − y − xex = 0.

Solution.




Example 2


Solution.

We will solve this problem by using the method of variation of a constant. First we find the general solution of the

homogeneous equation:








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Example 3

Solve the equation y' − 2y = x.

Solution.

A. First we solve this problem using an integrating factor. The given equation is already written in the standard form.

Therefore




Then

B. Now we construct the solution by the method of variation of a constant. Consider the corresponding homogeneous

equation:

and find its general solution.

where C again denotes any real number. Notice that at C = 0, we get y = 0 that is also a solution of the homogeneous

equation.

Next we suppose that C is a function of x and substitute the solution y = C(x)e2x into the initial nonhomogeneous

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equation. We can write

Hence,

This integral was already found above in section A, so we obtain

As a result, the general solution of the nonhomogeneous differential equation is given by

As you can see, both methods give the same answer :).

Example 4

Solve the differential equation x2y' + xy + 2 = 0.

Solution.

We solve this problem using the method of variation of a constant. For convenience, we write this equation in the

standard form:



After easy transformations we find the answer y = C/x, where C is any real number. The last expression includes the

case y = 0, which is also a solution of the homogeneous equation.

Now we replace the constant C with the function C(x) and substitute the solution y = C(x)/x into the initial

nonhomogeneous differential equation. Since

we obtain

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Example 5

Solve the initial value problem:

Solution.


Here


We can take the function u(x) = cos x as the integrating factor. One can make sure that the left side of the equation is

the derivative of the product y(x)u(x):





Example 6

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Solve the differential equation (IVP) with the initial condition y(1) = 2.

Solution.


We can take the function u(x) = x3 as the integrating factor. One can check that the left side of

the equation is the derivative of the product y(x)u(x):


Now we can find the constant C using the initial condition y(1) = 2. Substituting the general solution

into this condition gives:


Example 7

Find the general solution of the differential equation y = (2y4 + 2x)y'.

Solution.

It can be seen that this equation is not linear with respect to the function y(x). However, we can try to

find the solution for the inverse function x(y). We write the given equation through differentials and

make some transformations:

Now we see that we have a linear differential equation with respect to the function x(y). We can solve it with help of

the integrating factor:


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Formation of differential equations.

Q. 1. Form the differential equation of the following family of curves:

a. y=

b. y = asin(x+b)

c. y = Acosmx + Bsinmx ,

d. y = c+cx+cx2

e. y = e2X

(a+bx)

f. ex (acosx +bsinx)

g. Y = ae3X

+ be-2x

h. (x-a)2 + (y-b)

2 = r

2 .

i. Family of ellipses having foci on x-axis and centre at the origin.

Q. 2. Family of circles touching x-axis at the origin.

Q.3. Family of parabolas having vertex at the origin and axis along positive direction of x-axis.

Q.4 Family of curves y = ae2x

+ be-x

Q.5. Family of circles having centre on the x- axis and the radius is unity.

Q.6. Family of circles in the first quadrant which touches the co-ordinate axis.

Variable Separable

Q.1. Solve

Q. 2. Solve .

Q.3. Solve

Q. 4. Solve sin -1 = (x+y)

Q. 5. Solve cos -1 = (x+y)

Q. 6 . log =3x+4y, given y =0 when x = 0.

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Q.7. Solve

Q. 8 . Solve .

Q. 4 . Prove that is homogeneous and solve it.

Q. 5 Solve (x+y)dy + (x-y)dx = 0 given y = 1 when x = 1.

Q. 6 Solve (x2-y

2)dx+2xydy=0 given y = 1 when x = 1.

Q. 7 Solve (x2+xy)dy = (x

2+y

2)dx.

Q. 8 Solve (3xy+y2)dx + (x2+xy)dy=0.

Q. 9 Solve (xdy-ydx) = dx.

Q. 10 Solve (xdy-ydx)y =(ydx+xdy)x .

Q 11 .

Linear first order type

Q. 1 Solve

Q. 2. Solve

Q. 3. dx + xdy =

Q. 4 .Solve

Q. 5 . Solve xdy =

Q. 6. Solve

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Q. 7.

Q. 8.

Q. 9 .

Q. 10.

Q. 11.

Q. 12 (1+x2)dy+2xydx=cotxdx.

Q. 13 . ; y = 0 when x = 1.

Q. 14

Q. 15. given y = 2 when x = p/2.

Q.16 . (tan-1y-x)dy=(1+y2)dx.

Q. 17 .

Q.18 . , given y = 0 when x = 0.

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