Question

Solve: dy/dx + ylogy/x = y(logy)^2/x^2

Answer 1

$\frac{dy}{dx}+\frac{y\log{y}}{x}=\frac{y{\log^2{y}}}{x^2}$

we will seek solution in the form $y=10^{kx}$ , $\frac{dy}{dx}={k}\ln{10}*10^{kx}$

${k}\ln{10}*10^{kx}+10^{kx}\frac{kx}{x}=10^{kx}\frac{(kx)^2}{x^2}$

${k}{\ln10}+k=k^2$

$k(k-(1+\ln{10})=0$

$k_1=0$ ; $k_2=1+{1}\ln{10}$

$y_1=10^{0*x}=1$

$y_2=10^{(1+\ln{10})x}$