# Homework Help: (Ordinary) Differential Equation Trouble

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1. Feb 15, 2017

### Euler2718

1. The problem statement, all variables and given/known data

Find the solution of the differential equation by using appropriate method:

$$t^{2}y^{\prime} + 2ty - y^{3} = 0$$

2. Relevant equations

I'm thinking substitution method of a Bernoulli equation: $v = y^{1-n}$

3. The attempt at a solution

$$t^{2}y^{\prime} + 2ty - y^{3} = 0$$
$$\implies \frac{t^{2}y^{\prime}}{y^{3}} + \frac{2t}{y^{2}} = 1$$
Let $v = \frac{1}{y^{2}}$, then $v^{\prime} = -\frac{2y^{\prime}}{y^{3}}$

This is where I'm a little lost with respect to the given solution. It tells me that after substituing in $v$ and $v^{\prime}$ I should be getting:
$$t^{2}v^{\prime} - 2tv = -2$$
But I get:
$$-\frac{1}{2}t^{2}v^{\prime} + 2tv =1$$

Is there some algebra I'm messing up (or flat out not seeing)? Or is this not the best way to approach the problem? This is an elementary level ODE course so it shouldn't require anything too advance.

2. Feb 15, 2017

### LCKurtz

Your work looks correct. One way to know for sure is to push it through to a solution, then check whether your solution satisfies the original DE.

3. Feb 16, 2017

### Euler2718

Is this :

$$v^{\prime} - \frac{4}{t}v = \frac{2}{t^{2}}$$

Valid? Can I divide through by $t^{2}$ to make it into the form: $y^{\prime}(t) + p(t)y(t) = q(t)$. I still can't seem to get an answer that works if I substitute it back in.

4. Feb 16, 2017

### LCKurtz

Aren't you missing a sign on the right side from what you posted earlier (above)?

5. Feb 16, 2017

### Euler2718

Oh, that should be a $-\frac{2}{t^{2}}$. But that was just a LaTeX mistake.

6. Feb 16, 2017

### LCKurtz

In that case, you need to show the rest of your work before we can help you find what is wrong.

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