• September 22nd 2008, 02:41 AM
hats_06
I've got a few of these separable differential questions to ask. The thing I'm most confused about is what to do with the constant when you are given what y and x equal.

Solve each of the initial value problems-

1) e^x.y.(dy/dx) = 1 y = 0 when x = 0

2) xy.(dy/dx) = 1 + y^2 y = 0 when x = 2

Find the particular solution-

3) dy/dx = (1/3) + y^2 y = 1/2 when x = pi/4

4) dh/dt = 2h(3-h) where h = 1 when t = 0

Also these ones.

Solve:

5) (x^2 + 1).(dy/dx) + y^2 + 1 = 0

Sorry about all the questions, but im really confused with these (Crying)
• September 22nd 2008, 03:00 AM
mr fantastic
Quote:

Originally Posted by hats_06
I've got a few of these separable differential questions to ask. The thing I'm most confused about is what to do with the constant when you are given what y and x equal.

Solve each of the initial value problems-

1) e^x.y.(dy/dx) = 1 y = 0 when x = 0

2) xy.(dy/dx) = 1 + y^2 y = 0 when x = 2

Find the particular solution-

3) dy/dx = (1/3) + y^2 y = 1/2 when x = pi/4

4) dh/dt = 2h(3-h) where h = 1 when t = 0

Also these ones.

Solve:

5) (x^2 + 1).(dy/dx) + y^2 + 1 = 0

Sorry about all the questions, but im really confused with these (Crying)

How much experience do you actually have in solving differential equations? I'll do one.

2) xy.(dy/dx) = 1 + y^2 y = 0 when x = 2.

$\frac{dy}{dx} = \frac{1 + y^2}{xy} \Rightarrow \frac{y}{1 + y^2} \, dy = \frac{1}{x} \, dx$

$\Rightarrow \int \frac{y}{1 + y^2} \, dy = \int \frac{1}{x} \, dx$

$\Rightarrow \frac{1}{2} \ln | 1 + y^2 | = \ln |x | + C$.

Substitute y = 0 when x = 2: $\frac{1}{2} \ln | 1 | = \ln | 2 | + C \Rightarrow C = - \ln 2$.

Therefore:

$\frac{1}{2} \ln | 1 + y^2 | = \ln |x | - \ln 2$

$\Rightarrow \ln \sqrt{1 + y^2} = \ln \left| \frac{x}{2} \right|$

$\Rightarrow \sqrt{1 + y^2} = \frac{x}{2}$.

Therefore y = .....
• September 22nd 2008, 03:07 AM
hats_06
Thanks for that!