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:

$$y^4=t^4\left(\ln\left(t^4 \right)+C \right)$$
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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