Question

y'-x^4y =x^4 e^2/5x^5 ​

Answer 1

$y'-x^4y=x^4e^{dfrac{2}{5}x^5}\ dfrac{y'}{e^{frac{x^5}{5}}}-dfrac{x^4y}{e^{frac{x^5}{5}}}=x^4e^{frac{x^5}{5}}\ (*);;;;left(e^frac{-x^5}{5}right)'=e^frac{-x^5}{5}cdot(-x^4)\ y'*e^frac{-x^5}{5}right+y*e^frac{-x^5}{5}cdot(-x^4)=x^4e^{frac{x^5}{5}}\ left(ye^frac{-x^5}{5}right)'=x^4e^{frac{x^5}{5}}\ ye^frac{-x^5}{5}=int x^4e^{frac{x^5}{5}}dx$

$(**);;;int x^4e^{frac{x^5}{5}}dx=int e^{frac{x^5}{5}}d(frac{x^5}{5})=e^{frac{x^5}{5}}+C_1\ ye^frac{-x^5}{5}=e^{frac{x^5}{5}}+C_1\ y=e^{frac{2x^5}{5}}+C_1e^{frac{x^5}{5}}$

Answer 2

Спасибо