Applied mathematics (bernoulli equation)

Question

Can everybody solve this equation? $$xy-y'=y^4e^{-3x^2/2}$$ Can you please what "the material" that explain this equation. I saw in YouTube is "Bernoulli equation", is it right?

Answer 1

$$xy-y'=y^4e^{-3x^2/2}$$ Is indeed a Bernouilli's equation

Divide by $y^4 (y \ne 0)$ $$\frac x {y^3}-\frac {y'}{y^4}=e^{-3x^2/2}$$ Note that $(\frac 1 {y^3})'=-\frac {3y'}{y^4 }$ $$\frac x {y^3}+\frac 13 \left(\frac {1}{y^3}\right)'=e^{-3x^2/2}$$ Substitute $v=\frac 1 {y^3}$ $$v'+3vx=3e^{-3x^2/2}$$

This last equation is linear of first order. You can easily integrate it..