MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS Section 2.6 (ODE Book)
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) = 2xy2 + 2y and N (x, y) = 2x2y + 2x. Hence My(x, y) = 4xy + 2, and Nx(x, y) = 4xy + 2.
Therefore, the differential equation is exact. Therefore, we have
(x, y) = 2xy2 + 2ydx = x2y2 + 2xy + c(y).
Moreover, 2x2y + 2x = N (x, y) = y(x, y) = 2x2y + 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
x2y2 + 2xy = k,
for some constant k. 6. First the equation is transformed into
(ax - by)
+ (bx
-
cy)
dy dx
=
0.
Therefore, we have My(x, y) = -b and Nx(x, y) = b and hence the equation is not exact.
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 (x, y) = ex sin y - 2y sin xdx = ex sin y + 2y cos x + c(y).
1
2
MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS
Moreover, we must have
ex cos y + 2 cos x = N (x, y) = y(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 + (x2ex cos y)dy = 0.
We have
(x, y) = x2ex cos ydy = x2ex sin y + c(x).
Moreover, we must have xex(x + 2) sin y = M (x, y) = x(x, y) = sin y(2xex + x2ex) + 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
x2ex sin y = k,
where k is some constant.
26. We must have
?(x, y)(-e2x - y + 1) + ?(x, y)y = 0.
If this is to be an exact equation, we should have
?y(-e2x - y + 1) - ? = ?x. If ? is a function in x alone, then the above become
-? = ?x
and we see that we may take
? = e-x.
Now upon multiplying both sides by this integrating factor, we get
(-ex - e-xy + e-x) + e-xy = 0.
We have
(x, y) = e-xdy = e-xy + c(x).
Moreover, we must have
-e-xy + c(x) = -ex - e-xy + e-x.
This gives that
c(x) = e-x - ex, and hence c(x) = -e-x - ex + c.
MAS113 CALCULUS II, TUTORIAL 11 SOLUTIONS
3
Therefore, the solution is
e-x(y - 1) - ex = k, or y = kex + e2x + 1.
28. We must have
?(x, y)y + ?(x, y)(2xy - e-2y)y = 0
and ?yy + ? = ?x(2xy - e-2y) + 2y?.
If we assume that ? is a function of y alone, then the above becomes
?yy = ?(2y - 1).
Solving this differential equation in ?, we get that ?(x, y) = 1 exp (2y) . y
Now the differential equation becomes
e2y + 2xe2y - 1 y = 0. y
We have
(x, y) = e2ydx = xe2y + c(y).
Moreover, we also have
2xe2y - 1 = 2xe2y + c(y). y
This gives that
c(y) = - 1 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 y2. We get
(4x3 + 3y) + (3x + 4y2)y = 0
which is an exact equation! Therefore, the integrating factor is ? = y2. Also, we have
(x, y) = (4x3 + 3y)dx = x4 + 3xy + c(y).
Furthermore, we get
3x + 4y2 = 3x + c(y).
It must be that c(y) = 4y2 and hence
c(y)
=
4 3
y3
+
c.
The solution to the differential equation is
x4
+
3xy
+
4 3
y3
=
k.
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