# owen b's question at Yahoo! Answers regarding a first order homogenous ODE

#### MarkFL

Staff member
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

Staff member
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:

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