# Thread: First Order Differential Equation

1. ## First Order Differential Equation

find the general solution of the first order differential equation, t2 y’ – 2 ty = 3

2. Originally Posted by winsome
find the general solution of the first order differential equation, t2 y’ – 2 ty = 3
I'm assuming you mean $\displaystyle t^2y^{\prime}-2ty=3$.

First note that this is the same thing as saying $\displaystyle y^{\prime}-\frac{2}{t}y=\frac{3}{t^2}$.

Now this is a first order linear DE. Do you know how to proceed using the integrating factor method?

3. i got integrating factor of this equaiton is t^-2
is that is right ?
if yes then i m unable to get its final answer

4. Yes it's right.

So multiply both sides by the integrating factor to get

$\displaystyle \displaystyle t^{-2}\frac{dy}{dt} - 2t^{-3}\,y = 3t^{-4}$

$\displaystyle \displaystyle \frac{d}{dt}(t^{-2}y) = 3t^{-4}$.

You should be able to go from here.

5. thanks

6. Originally Posted by Prove It
Yes it's right.

So multiply both sides by the integrating factor to get

$\displaystyle \displaystyle t^{-2}\frac{dy}{dt} - 2t^{-3}\,y = 3t^{-4}$

$\displaystyle \displaystyle \frac{d}{dt}(t^{-2}y) = 3t^{-4}$.

You should be able to go from here.

thanks for providing solution