# Homework Help: Solving a DE via Substitution

1. Sep 26, 2011

### Lancelot59

I need to find a 1 parameter family of solutions to:

$$\frac{dy}{dt}=-\frac{1}{t^{2}} - \frac{y}{t}+y^{2}$$

By making the substitution:
$$y=\frac{1}{t}+u$$
and then reducing it to a Bernoulli equation in u.

I first took the derivative of the substitution.
$$\frac{dy}{dt}=\frac{1}{t^{2}}+\frac{du}{dt}$$

Then substituted:
$$\frac{1}{t^{2}}+\frac{du}{dt}=-\frac{1}{t^{2}} - \frac{\frac{1}{t}+u}{t}+(\frac{1}{t}+u)^{2}$$

After some reduction I eventually got to:
$$\frac{1}{t^{2}}+\frac{du}{dt}=\frac{u+u^{2}t}{t}-1$$
I heard we were supposed to get something separable out of this, but that's not what I have here. What do I do next?

2. Sep 26, 2011

### LCKurtz

The derivative of 1/t is -1/t2. If you are careful with your algebra it should indeed simplify to a Bernoulli equation.

3. Sep 26, 2011

### Lancelot59

I now have:
$$u^{2}+\frac{u}{t}+\frac{1}{t^2}-\frac{du}{dt}=1$$

4. Sep 26, 2011

### LCKurtz

Close, but check your algebra. I got

$$u' -\frac 1 t u = u^2$$

Those 1/t2 terms should cancel out. I'm guessing it has something to do with that extra 1 you have too.

5. Sep 26, 2011

### Lancelot59

The 1 comes from reducing
$$\frac{y}{t}$$

$$\frac{\frac{1}{t}+u}{t}$$

6. Sep 26, 2011

### LCKurtz

Like I said before, the non-cancelled 1/t2 and the extraneous 1 are probably related. If you are now taking differential equations, you should be able to correctly simplify this expression. You don't get a 1.

7. Sep 26, 2011

### Lancelot59

I keep getting it for some reason :(

Nevermind, I'm stupid. Lemme keep working.

So now I have the right form. Now what? I'm reading the book and it isn't making sense to me right now.

Last edited: Sep 26, 2011
8. Sep 26, 2011

### process91

What part are you at now?

(I solved this along with you)

9. Sep 26, 2011

### LCKurtz

If you have the above equation, multiply both sides by u-2 and make the substitution w = u-1.

10. Sep 26, 2011

### Lancelot59

So I then end up with
$$\frac{w}{t}-w\frac{du}{dt}=1$$

11. Sep 26, 2011

### process91

I think you need to check that again. Also, using the substitution, what is w'?

12. Sep 26, 2011

### process91

13. Sep 26, 2011

### Lancelot59

I entered it backwards, my mistake.
Doing the substitution:
$$w=u^{-1}$$
$$\frac{dw}{dt}=\frac{1}{u^{2}}\frac{du}{dt}$$
$$u^{2}w'=\frac{du}{dt}$$

$$w'u-\frac{w}{t}=1$$
I'll start going through those notes now.

14. Sep 26, 2011

### process91

Hmm... still not quite right.

Starting with
$$u' -\frac 1 t u = u^2$$
Let $w=u^{-1}$, then $w'=-u^{-2} u'$. This serves as the motivation for the next step: we want to get a linear differential equation in terms of w, so only w and w' should be present. To do so, let's multiply both sides of the equation by $-u^{-2}$, in doing so we will get w' on the left, which is very desirable.

$$-u^{-2}u' + \frac 1 t u^{-1} = -1$$

Now we can make some nice substitutions:

$$w' + \frac 1 t w = -1$$

You should be able to take it from here.

15. Sep 26, 2011

### Lancelot59

I really should, but I can't. I'm not having the easiest time with this class.

16. Sep 26, 2011

### process91

What is the trouble you're having from the last step? Have you covered how to solve a first-order linear homogeneous differential equation?

17. Sep 26, 2011

### process91

To clarify some possible confusion, at this point you would simply solve for w. This will give you an equation for w in terms of x, and then you can substitute $u^{-1}=w$.

18. Sep 26, 2011

### Lancelot59

We have covered that. I'm trying to figure out variation of parameters, since it isn't seperable. I'll have a chance to talk to my professor tomorrow, and I'll come back if I need more help. Thank you both for all the help you've given me.