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

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    First order Ode
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The discussion focuses on solving the first-order homogeneous ordinary differential equation (ODE) given by dy/dt = t^3/y^3 + y/t. The approach involves recognizing it as a Bernoulli equation and applying the substitution u = y/t, which simplifies the equation. After substituting, the ODE is transformed into a separable form, allowing for integration. The integration leads to an implicit solution expressed as y^4 = t^4(ln(t^4) + C). This method effectively demonstrates the process of solving the given ODE.
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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.
 
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Hello owen b,

We are given to solve:

$$\frac{dy}{dt}=\frac{t^3}{y^3}+\frac{y}{t}$$

I would first express the ODE as:

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

Now, use the substitution:

$$u=\frac{y}{t}\implies y=ut\implies\frac{dy}{dt}=u+\frac{du}{dt}t$$

And our ODE become:

$$u+\frac{du}{dt}t=u^{-3}+u$$

$$\frac{du}{dt}t=u^{-3}$$

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

$$\int u^3\,du=\int\frac{dt}{t}$$

$$\frac{u^4}{4}=\ln|t|+C$$

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

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

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