Show that the equation is not exact, but becomes exact when multiplied by the given integrating factor. Then solve the equation.
$y'=e^{2x}+y−1$ with integrating factor $\mu(x,y)=1/xy^3$.
I got that $x+[(1+y^2)/y^3]y'=0$... How do you solve from here?
On
Simply integrate the equation, $$x+[(1+y^2)/y^3]y'=0$$ $$\int \dfrac {(1+y^2)}{y^3}dy=-\int x dx$$ $$\int \dfrac {dy}{y^3}+\int \dfrac {dy}{y}=-x^2/2+K$$ Another way to integrate the differential equation $$y'=e^{2x}+y−1$$ Substitute $z=e^{2x}+y−1$ and $z'=y'+2e^{2x}$
You get $$z'-z=2e^{2x}$$ $$ze^{-x}= 2\int e^{x}dx$$ $$z=2 e^{2x}+Ke^{x}$$ $$\boxed{y=1+ e^{2x}+Ke^{x}}$$
write your equation in the form $$y'-y=e^{2x}-1$$ and $$\mu=e^{\int-1dx}=e^{-x}$$ and then $$\frac{\frac{dy}{dx}}{e^x}-e^{-x}y=\frac{e^{-2x}-1}{e^x}$$ then $$\frac{d}{dx}\left(\frac{y(x)}{e^x}\right)=\frac{e^{2x}-1}{e^x}$$ can you finish? the solution is given by $$y(x)=c_1 e^x+e^x \left(x+e^{-x}\right)$$