Question

xy'+y+x^{2}y^{2}e^{x}=0

Answer 1

\[xy'+y+x^{2}y^{2}e^{x}=0\]

Answer 2

Bernoulli Equation
\[x\frac{dy}{dx}+e^{x}x^{2}y^{2}+y=0\]\[x\frac{dy}{dx}+y=-e^{x}x^{2}y^{2}\]\[-\frac{1}{xy^{2}}x\frac{dy}{dx}-\frac{1}{xy}=e^{x}\]
\[v=\frac{1}{y}\]\[\frac{dv}{dx}=-\frac{1}{y^2}\frac{dy}{dx}\]
\[\frac{dv}{dx}=\frac{v}{x}=e^{x}x\]
\[\mu=e^{\int-\frac{1}{x}dx}=\frac{1}{x}\]
\[\frac{1}{x}\frac{dv}{dx}-\frac{v}{x^2}=e^{x}\]
\[-\frac{1}{x^2}=\frac{d}{dx}\left(\frac{1}{x}\right)\]
\[\frac{1}{x}\frac{dv}{dx}+\frac{d}{dx}\left(\frac{1}{x}\right)v=e^{x}\]\[\frac{d}{dx\left(\frac{v}{x}\right)}=e^{x}\]\[\frac{v}{x}=e^{x}+c_1\]\[v=x(e^{x}+c_1)\]\[y=\frac{1}{x(e^{x}+c_1)}\]pz

Answer 3

i knew it was non linear , thankyou

Answer 4

im confused on your third line did you divide the second line by \[-1/(x^{2}y^{2})\]