# Solution for these Differential equations

1. Oct 30, 2005

### Reshma

Hi everyone. I'm trying to solve these differential equations but I could not crack a single nut. I seem to have lost my memory on solving differential equations . Please help me refresh it by providing useful hints.
I'm unable to separate the variables in the following. Perhaps I'm missing out on something important.

1] $$\frac{dy}{dx} + 2y = y^2e^{2x}$$

2] $$2y\frac{dy}{dx} + y^2 = \frac{x}{2}e^{-x}$$

3] $$x^2\frac{dy}{dx} - 2xy = \frac{1}{x}$$

2. Oct 30, 2005

### saltydog

First one is a Riccati. Make the change of variables usually done for such equations and see what happens.

3. Oct 30, 2005

### Physics Monkey

For the second one, can you simplfy the derivative term? After a well chosen substitution, the differential equation becomes linear.

For the third one, the equation is linear first order and there is a general method available.

4. Oct 31, 2005

### Reshma

Thank you so much for the help.

Well, I was able to solve the third one!

Bringing the equation in the general form:
$$\frac{dy}{dx} + P(x)y = Q(x)$$

$$\frac{dy}{dx} - \frac{2}{x} y = \frac{1}{x^3}$$

Setting $$y = u(x)v(x)$$

So,
$$\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}$$

$$u\left(\frac{dv}{dx} - \frac{2}{x}v\right) + v\frac{du}{dx} = \frac{1}{x^3}$$.....(1)

Solving for v:
$$\frac{dv}{dx} - \frac{2}{x} v = 0$$

On solving:
$$\ln v = \ln x^2$$
$$v = x^2$$

Again on substituting for v in (1):
$$u = -\frac{1}{4x^4} + C$$

General formula:
$$y = v(x)\int \frac{Q(x)}{v(x)} dx + (C)v(x)$$

$$y = -\frac{1}{4x^2} + C{x}^2$$
Hope I'm right.
Sorry I could not find any suitable substitution for the second one . Please help!

Last edited: Oct 31, 2005
5. Oct 31, 2005

### Reshma

Ricatti? I haven't studied any differential equation like that. How do you solve such equations.

6. Oct 31, 2005

### saltydog

First place it into standard form:

$$y^{'}+Q(x)y+R(x)y^2=P(x)$$

Now, make the transformation:

$$y=\frac{u^{'}}{Ru}$$

Can you now substitute this into the ODE? I'll start it for you:

$$y^{'}=\frac{Ruu^{''}-u^{'}(Ru^{'}+uR^{'})}{(Ru)^2}$$

right?

Make the other ones to get:

$$\frac{Ruu^{''}-u^{'}(Ru^{'}+uR^{'})}{(Ru)^2}+\frac{Qu^{'}}{Ru}+R\left(\frac{u^{'}}{Ru}\right)^2=P$$

Now simplify and obtain a second order in u. Solve, convert back to y, and I want a plot.

Edit: Suppose that last one looks a bit intimidating. That's just the general expression though. For your equation a lot of stuff just drops out leaving a simple second order to solve. Try it.

Last edited: Oct 31, 2005
7. Oct 31, 2005

### Physics Monkey

For the second one, focus on the term $$2 y \frac{dy}{dx}$$, can you write this as something more convenient? Hint: notice that the only other y term you have is $$y^2$$.

8. Nov 1, 2005

### Reshma

I had tried it:

Set,
$$u = y^2$$

So,
$$\frac{du}{dx} = 2y\left(\frac{dy}{dx}\right)$$

So the eqaution becomes,
$$\frac{du}{dx} + u = \frac{x}{2} e^{-x}$$

But, I'm still unable to separate the variables. Should I adopt a different method?

9. Nov 1, 2005

### Benny

I don't think you can separate variables. You could try the integrating factor technique since you have a first order linear differential equation.

10. Nov 1, 2005

### saltydog

Reshma, rearrange the equation to:

$$2ydy+y^2dx=\frac{x}{2}e^{-x}dx$$

or:

$$\left(\frac{x}{2}e^{-x}-y^2\right)dx-2ydy=0$$

Now, we can make this exact right? You know, the partial of M with respect to y, partial of N with respect to x, do that arithmetic, get some function of x or y, then e to the integral of that function is the integrating factor right? You know this makes two plots now.

Last edited: Nov 1, 2005
11. Nov 1, 2005

### Physics Monkey

What Salty suggested is a nice way of proceeding, or as Benny said, you now have a linear first order equation in u and a general approach exists as I said before.

12. Nov 2, 2005

### Reshma

Wow, thanks for your help, Saltydog and PhysicsMonkey. I got all the solutions!!