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.

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download