lesson 6: exact equations determine whether or not each of
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Lesson 6: Exact equations
Determine whether or not each of equations below are exact. If it is exact findthe solution by two methods.
1. (2x + 4y) + (2x− 2y)y′ = 0
Solution. The differential equation is not exact since
(2x + 4y)′y = 4,
while(2x− 2y)′x = 2.
2. (3x2 − 2xy + 2)dx + (6y2 − x2 + 3)dy = 0
Solution. The differential equation is exact since
(3x2 − 2xy + 2)′y = −2x,
(6y2 − x2 + 3)′x = −2x.
Let us find a solution.
Method 1. Integrating the first equation, we have∫(3x2 − 2xy + 2)dx = x3 − x2y + 2x + h(y).
To find h(y) we have−x2 + h′(y) = 6y2 − x2 + 3.
Thereforeh′(y) = 6y2 + 3,
and
h(y) =∫
(6y2 + 3)dy = 2y3 + 3y.
Therefore, the solution is
x3 − x2y + 2x + 2y3 + 3y = c.
Method 2. By integrating on curve (0,0) (x, y) we have∫ x
0
(3u2 − 2u× 0 + 2)du +∫ y
0
(6v2 − x2 + 3)dv
= x3 + 2x + 2y3 − x2y + 3y = c.
3. (2xy2 + 2y) + (2x2y + 2x)y′ = 0
Solution.(2xy2 + 2y)′y = 4xy + 2,
(2x2y + 2x)′x = 4xy + 2.
That is the differential equation is exact. Find a solution by two methods.
Method 1. Integrating the first equation we have∫(2xy2 + 2y)dx = x2y2 + 2xy + h(y).
1
2
Let us find h(y). We have:
2x2y + 2x + h′(y) = 2x2y + 2x.
That ish′(y) = 0,
and the solution isx2y2 + 2xy = c.
Method 2. By integrating on the curve we have∫ x
0
(2u× 02 + 2× 0)du +∫ y
0
(2x2v + 2x)dv = x2y2 + 2xy = c.
4.dy
dx= −ax + by
bx + cy.
Solution. Rewriteax + by
bx + cy+ y′ = 0.
Then, (ax + by
bx + cy
)′x6= 0
in general, while(1)′y = 0.
That is the differential equation is not exact. With what c it is exact?5. (ex sin y − 2y sin x)dx + (ex cos y + 2 cos x)dy = 0
Solution.(ex sin y − 2y sin x)′y = ex cos y − 2 sinx,
(ex cos y + 2 cos x)′x = ex cos y − 2 sinx.
Therefore, the differential equation is exact.
Method 1. ∫(ex sin y − 2y sinx)dx = ex sin y + 2y cosx + h(y).
Next,ex cos y + 2 cos x + h′(y) = ex cos y + 2 cos x.
Therefore,h′(y) = 0,
andex sin y + 2y cosx = c
is a solution.
Method 2.∫ x
0
(eu × sin 0− 2× 0 sin u)du +∫ y
0
(ex cos v + 2 cos x)dv = ex sin y + 2y cos x = c.
6. (ex sin y + 3y)dx− (3x− ex sin y)dy = 0.
3
Solution.(ex sin y + 3y)′y = ex cos y + 3
−(3x− ex sin y)′x = −3 + ex sin y
That is the differential equation is not exact.
7. (yexy cos 2x− 2exy sin 2x + 2x)dx + (xexy cos 2x− 3)dy = 0.
Solution.
(yexy cos 2x− 2exy sin 2x + 2x)′y = exy cos 2x + xyexy cos 2x− 2xexy sin 2x.
(xexy cos 2x− 3)′x = exy cos 2x + xyexy cos 2x− 2xexy sin 2x.
Thus, the differential equation is exact.Then, ∫
(xexy cos 2x− 3)dy = cos 2xexy − 3y + h(x)
By differentiating in x we have
−2 sin 2xexy + y cos 2xexy + h′(x) = yexy cos 2x− 2exy sin 2x + 2x
Therefore,h(x) = x2,
and the solution iscos 2xexy − 3y + x2 = c.
8. (y/x + 6x)dx + (log x− 2)dy = 0, x > 0.Solution.
(y/x + 6x)′y = 1/x.
(log x− 2)′x = 1/x.
The differential equation is exact.Therefore (x is positive),∫
(y/x + 6x) = y log x + 6x2 + h(y).
Differentiation in y yields
log x + h′(y) = log x− 2,
h(y) = −2y.
The solution isy log x + 6x2 − 2y = c.
9. (x log y + xy)dx + (y log x + xy)dy = 0, x > 0, y > 0.
Solution.(x log y + xy)′y =
x
y+ x
(y log x + xy)′x =y
x+ y
The differential equation is not exact.10.
xdx
(x2 + y2)3/2+
ydy
(x2 + y2)3/2= 0.
4
Solution.( x
(x2 + y2)3/2
)′y
= −3yx(x2 + y2)1/2
(x2 + y2)3= − 3yx
(x2 + y2)5/2.
( y
(x2 + y2)3/2
)′x
= − 3yx
(x2 + y2)5/2.
Therefore, the differential equation is exact.Then ∫
xdx
(x2 + y2)3/2=
12
∫d(x2 + y2)
(x2 + y2)3/2
= − 1(x2 + y2)1/2
+ h(y)
Next,y
(x2 + y2)3/2+ h′(y) =
y
(x2 + y2)3/2.
Therefore, the solution is1
(x2 + y2)1/2= c.
Find an integrating factor and solve the following equations11. (3x2y + 2xy + y3)dx + (x2 + y2)dy = 0.
Solution. We have:
(3x2y + 2xy + y3)′y = 3x2 + 2x + 3y2,
(x2 + y2)′x = 2x.
Therefore,M ′
y(x, y)−N ′x(x, y)
N(x, y)= 3.
Thus, an integrating factor is
I(x, y) = I(x) = exp(∫
3dx)
= e3x.
Then, the differential equation
e3x(3x2y + 2xy + y3)dx + e3x(x2 + y2)dy = 0
is exact.The solution of the differential equation is
∫ y
0
e3x(x2 + v2)dv = e3x(x2y +
y3
3
).
12.y′ = e2x + y − 1.
Solution. Let us write the equation in the form
(e2x + y − 1)dx− dy = 0.
We have(e2x + y − 1)′y = 1
(−1)′x = 0.
5
Therefore,M ′
y(x, y)−N ′x(x, y)
N(x, y)= −1.
Then, an integrating factor is
I(x, y) = I(x) = e−x,
and the differential equation
(ex + e−xy − e−x)dx− e−xdy = 0.
is exact.
13. dx + (x/y − sin y)dy = 0.
Solution. We have(1)′y = 0,
(x/y − sin y)′x =1y.
ThereforeN ′
x(x, y)−M ′y(x, y)
M(x, y)=
1y.
Then, an integrating factor is
I(x, y) = I(y) = exp(∫
dy
y
)= y,
and the differential equation
ydx + (x− y sin y)dy = 0
is exact.
14. ydx + (2xy − e−2y)dy = 0.
Solution.(y)′y = 1,
(2xy − e−2y)′x = 2y
ThereforeN ′
x(x, y)−M ′y(x, y)
M(x, y)=
2y − 1y
.
Then, an integrating factor
I(x, y) =∫
2y − 1y
dy = 2y − log |y|,and the differential equation
(2y2 − y log |y|)dx + (4xy2 − 2ye−2y + 2xy log |y| − 2y log |y|e−2y)dy = 0
is exact.
15. exdx + (ex cot y + 2y sec y)dy = 0.
Solution.(ex)′y = 0
6
(ex cot y + 2y sec y)′x = ex cot y
ThereforeN ′
x(x, y)−M ′y(x, y)
M(x, y)= cot y,
and the differential equation
cot yexdx + (ex cot2 y + 2y csc y)dy = 0
is exact.
16. [4(x3/y2) + (3/y)]dx + [3(x/y2) + 4y]dy = 0.
Solution.[4(x3/y2) + (3/y)]′y = −8x3/y3 − 3/y2
[3(x/y2) + 4y]′x = 3/y2
ThereforeN ′
x(x, y)−M ′y(x, y)
M(x, y)=
8x3/y3 + 6/y2
4x3/y2 + 3/y= 2y,
and the differential equation
[8(x3/y) + 6]dx + [6(x/y) + 8y2]dy = 0
is exact.E-mail address: [email protected]