# What is a solution to the differential equation #y'=xe^(x^2-lny^2)#?

we can separate it out first

.....spotting the pattern on the RHS

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To solve the differential equation (y' = xe^{x^2 - \ln(y^2)}), we start by simplifying and using separation of variables.

Given:

[y' = xe^{x^2 - \ln(y^2)}]

Notice that (e^{-\ln(y^2)} = e^{\ln(1/y^2)} = \frac{1}{y^2}). Thus, the equation becomes:

[y' = xe^{x^2}\frac{1}{y^2}]

This gives:

[y^2 dy = xe^{x^2} dx]

Now, integrate both sides:

[\int y^2 dy = \int xe^{x^2} dx]

The left side becomes:

[\frac{y^3}{3} = \frac{e^{x^2}}{2} + C]

So, the solution to the differential equation in implicit form is:

[\frac{y^3}{3} = \frac{e^{x^2}}{2} + C]

Where (C) is the constant of integration.

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