Question

Solve the differential equation :$y\log y\frac{dx}{dy}+x-\log y=0$.

Answer 1

We have,

$y\log y\frac{dx}{dy}+x-\log y=0$

Now,

$\frac{dx}{dy}+\frac{x}{y\log y}-\frac{\log y}{y\log y}=0$

$\frac{dx}{dy}+\frac{x}{y\log y}=\frac{1}{y}$

On Comparing that,

$\frac{dx}{dy}+Px=Q$

$P=\frac{1}{y\log y},Q=\frac{1}{y}$

$I.F.$

$I.F.=e^{\int Pdy}=e^{\int \frac{1}{y\log y}dy}=e^{\log \left(\log y\right)}=\log y$

Then,

$x\times I.F.=\int Q\times I.F.+C$

$x\log y=\int \frac{1}{y}\times \log ydy+C$

$x\log y=\frac{(\log y{)}^2}{2}+C$

Hence, this is the answer.