Question

𝑦′ − 𝑦 = 𝑒2𝑥𝑦3

Answer 1

$$y'-y = e^{2x}y^3$$

Substitute $v=y^{1-3}$

$$v'=-2y^{-3}y'$$

$$-\dfrac{1}{2}v'-v=e^{2x}$$

$$v'+2v=-2e^{2x}$$

Integrating factor

$$\mu(x)=e^{\int(2)dx}=e^{2x}$$

$$e^{2x}v'+2e^{2x}v=-2e^{4x}$$

$$d(e^{2x}v)=-2e^{4x}dx$$

Integrate

$$\int d (e^{2x}v)=-\int2e^{4x}dx$$

$$e^{2x}v=-\dfrac{1}{2}e^{4x}+C$$

$$v=-\dfrac{1}{2}e^{2x}+\dfrac{1}{2}Ce^{-2x}$$

Then

$$\dfrac{1}{y^2}=-\dfrac{1}{2}e^{2x}+\dfrac{1}{2}Ce^{-2x}$$

$$y^2=\dfrac{2}{Ce^{-2x}-e^{2x}}$$

$$y^2=\dfrac{2e^{2x}}{C-e^{4x}}$$