Owen b's question at Yahoo Answers regarding a first order homogenous ODE

[MarkFL]

Here is the question:

How to solve this equation? dy/dt= t^3/y^3 + y/t?


How to solve this equation? dy/dt= t^3/y^3 + y/t

what i understand is we have to use Bernoulli and then solve it using linear equation,is it?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello owen b,

We are given to solve:

\(\displaystyle \frac{dy}{dt}=\frac{t^3}{y^3}+\frac{y}{t}\)

I would first express the ODE as:

\(\displaystyle \frac{dy}{dt}=\left(\frac{y}{t} \right)^{-3}+\frac{y}{t}\)

Now, use the substitution:

\(\displaystyle u=\frac{y}{t}\implies y=ut\implies\frac{dy}{dt}=u+\frac{du}{dt}t\)

And our ODE become:

\(\displaystyle u+\frac{du}{dt}t=u^{-3}+u\)

\(\displaystyle \frac{du}{dt}t=u^{-3}\)

Separating variables and integrating (noting $t\ne0$), we obtain:

\(\displaystyle \int u^3\,du=\int\frac{dt}{t}\)

\(\displaystyle \frac{u^4}{4}=\ln|t|+C\)

\(\displaystyle u^4=\ln\left(t^4 \right)+C\)

Back-substituting for $u$, we obtain the implicit solution:

\(\displaystyle y^4=t^4\left(\ln\left(t^4 \right)+C \right)\)