MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS
Section 2.6 (ODE Book)
2. We have M (x, y) = 2x + 4y and N (x, y) = 2x − 2y. Hence
My (x, y) = 4, and Nx (x, y) = 2.
Therefore, the differential equation is not exact.
4. We have M (x, y) = 2xy 2 + 2y and N (x, y) = 2x2 y + 2x. Hence
My (x, y) = 4xy + 2, and Nx (x, y) = 4xy + 2.
Therefore, the differential equation is exact. Therefore, we have
Z
ψ(x, y) = 2xy 2 + 2ydx = x2 y 2 + 2xy + c(y).
Moreover,
2x2 y + 2x = N (x, y) = ψy (x, y) = 2x2 y + 2x + c′ (y).
We get that c′ (y) = 0 and hence c(y) = k for some constant. Therefore, the implicit solution
to the differential equation is
x2 y 2 + 2xy = k,
for some constant k.
6. First the equation is transformed into
dy
= 0.
dx
Therefore, we have My (x, y) = −b and Nx (x, y) = b and hence the equation is not exact.
(ax − by) + (bx − cy)
18. If the differential equation is separable, then M and N are functions of x and y only,
respectively. Therefore, we must have
My (x, y) = 0 and Nx (x, y) = 0.
Therefore, all separable differential equations are exact.
20. Upon multiplying both sides by the integrating factor, the equation becomes
(ex sin y − 2y sin x)xdx + (ex cos y + 2 cos x)dy = 0.
We have
ψ(x, y) =
Z
ex sin y − 2y sin xdx = ex sin y + 2y cos x + c(y).
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MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS
2
Moreover, we must have
ex cos y + 2 cos x = N (x, y) = ψy (x, y) = ex cos y + 2 cos x + c′ (y).
We must have that c′ (y) = 0 and hence c(y) = c for some constant c. Therefore, the implicit
form of the solution to the differential equation is
ex sin y + 2y cos x = k,
where k is some constant.
22. Upon multiplying both sides by the integrating factor, the equation becomes
(xex (x + 2) sin y)dx + (x2 ex cos y)dy = 0.
We have
ψ(x, y) =
Z
x2 ex cos ydy = x2 ex sin y + c(x).
Moreover, we must have
xex (x + 2) sin y = M (x, y) = ψx (x, y) = sin y(2xex + x2 ex ) + c′ (x).
We must have that c′ (x) = 0 and hence c(x) = c for some constant c. Therefore, the implicit
form of the solution to the differential equation is
x2 ex sin y = k,
where k is some constant.
26. We must have
µ(x, y)(−e2x − y + 1) + µ(x, y)y ′ = 0.
If this is to be an exact equation, we should have
µy (−e2x − y + 1) − µ = µx .
If µ is a function in x alone, then the above become
−µ = µx
and we see that we may take
µ = e−x .
Now upon multiplying both sides by this integrating factor, we get
(−ex − e−x y + e−x ) + e−x y ′ = 0.
We have
ψ(x, y) =
Z
e−x dy = e−x y + c(x).
Moreover, we must have
−e−x y + c′ (x) = −ex − e−x y + e−x .
This gives that
c′ (x) = e−x − ex , and hence c(x) = −e−x − ex + c.
MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS
Therefore, the solution is
e−x (y − 1) − ex = k, or y = kex + e2x + 1.
28. We must have
µ(x, y)y + µ(x, y)(2xy − e−2y )y ′ = 0
and
µy y + µ = µx (2xy − e−2y ) + 2yµ.
If we assume that µ is a function of y alone, then the above becomes
µy y = µ(2y − 1).
Solving this differential equation in µ, we get that
1
µ(x, y) = exp (2y) .
y
Now the differential equation becomes
¶
µ
1
2y
2y
y ′ = 0.
e + 2xe −
y
We have
Z
ψ(x, y) = e2y dx = xe2y + c(y).
Moreover, we also have
2xe2y −
This gives that
1
= 2xe2y + c′ (y).
y
1
c′ (y) = − and hence c(y) = − ln |y| + c.
y
Therefore, the solution is
xe2y − ln |y| = k.
30. Multiplying both side of the differential equation by y 2 . We get
(4x3 + 3y) + (3x + 4y 2 )y ′ = 0
which is an exact equation! Therefore, the integrating factor is µ = y 2 . Also, we have
Z
ψ(x, y) = (4x3 + 3y)dx = x4 + 3xy + c(y).
Furthermore, we get
3x + 4y 2 = 3x + c′ (y).
It must be that c′ (y) = 4y 2 and hence
4
c(y) = y 3 + c.
3
The solution to the differential equation is
4
x4 + 3xy + y 3 = k.
3
3
Section 11.1