Question

$\frac{dy}{dx}+\frac{y}{x}\ln y=\frac{y}{x^2}{\left(\ln y\right)}^2$
Solving it gives $x=\ln y(c{x}^2+\frac{1}{k})$
Find $5k$

Answer 1

$\frac{dy}{dx}+\frac{y}{x}\ln y=\frac{y}{x^2}{\left(\ln y\right)}^2$
Solving it gives $x=\ln y(c{x}^2+\frac{1}{k})-----1$
Find 5k?

Use: Linear DIFFERENTIAL form

Divide by $y(\ln y{)}^2$ to whole equation

$\frac{1}{y(\ln y{)}^2}\frac{dy}{dx}+\frac{1}{x\ln y}=\frac{1}{x^2}$

Put $\frac{1}{\ln y}=t$

$-\frac{1}{y(\ln y{)}^2}\frac{dy}{dx}=\frac{dt}{dx}$

$\frac{dt}{dx}-\frac{t}{x}=-\frac{1}{x^2}$

I.F$=e^{\int P\left(x\right)dx}=e^{\int -\frac{1}{x}dx}=e^{-\int \frac{1}{x}dx}=e^{\ln \frac{1}{x}}$

I.F$=\frac{1}{x}$

$y(I.F)=\int Q\left(x\right).(I.F)dx+C$

Here y variable is t

$\frac{t}{x}=\int \frac{-1}{x^2}\frac{1}{x}dx+C$

$\frac{1}{x\left(\ln y\right)}=\frac{1}{2x^2}+C$

$\frac{1}{x}=\ln y\left(\frac{1+2c{x}^2}{2x^2}\right)$

$x=\ln y(c^1{x}^2+\frac{1}{2})------2$

from $1$ and $2$

$k=2$

$5k=10$