Question

Solve xdy/dx +y = lnx.y^2

Answer 1

solve for what?

Answer 2

\[\begin{align*}x\frac{dy}{dx}+y&=\ln x\cdot y^2\\
x\frac{1}{y^2}\frac{dy}{dx}+\frac{1}{y}&=\ln x\end{align*}\]
Let \(t=y^{1-2}=\dfrac{1}{y}\), then \(\dfrac{dt}{dx}=-\dfrac{1}{y^2}\dfrac{dy}{dx}\):
\[\begin{align*}-x\frac{dt}{dx}+t&=\ln x\\
\frac{dt}{dx}-\frac{1}{x}t&=-\frac{\ln x}{x}\end{align*}\]
Find the integrating factor:
\[\mu(x)=\exp\left(\int-\frac{dx}{x}\right)=\exp(-\ln x)=-\frac{1}{x}\]
\[\begin{align*}-\frac{1}{x}\frac{dt}{dx}+\frac{1}{x^2}t&=\frac{\ln x}{x^2}\\
\frac{d}{dx}\left[-\frac{t}{x}\right]&=\frac{\ln x}{x^2}\\
-\frac{t}{x}&=\int\frac{\ln x}{x^2}~dx
\end{align*}\]
For the integral, integrate by parts:
\[\begin{matrix}f=\ln x&&&dg=\frac{dx}{x^2}\\df=\frac{dx}{x}&&&g=-\frac{1}{x}\end{matrix}\]
\[\int\frac{\ln x}{x^2}~dx=-\frac{\ln x}{x}+\int\frac{dx}{x^2}=-\frac{\ln x}{x}-\frac{1}{x}+C\]
So you get
\[\begin{align*}-\frac{t}{x}&=-\frac{\ln x}{x}-\frac{1}{x}+C\\\\
t&=\ln x+1-Cx
\end{align*}\]

Answer 3

Oh, and back-substituting again:
\[\frac{1}{y}=t\]