(Ordinary) Differential Equation Trouble

[Euler2718]

Homework Statement

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

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

Homework Equations

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

The Attempt at a Solution

[/B]
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.

 

[LCKurtz]

Homework Statement

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

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

Homework Equations

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

The Attempt at a Solution

[/B]
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.

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.

 

[Euler2718]

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.

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.

 

[LCKurtz]

But I get:
-\frac{1}{2}t^{2}v^{\prime} + 2tv =1

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.

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

 

[Euler2718]

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

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

 

[LCKurtz]

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

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