Question

solve:dz/dx+(z/x*log(z))=z/x^2 *(logz)^2

Answer 1

Solution:

$dz/dx+(z/x*log(z))=z/x^2 *(logz)^2\\\\ \frac{1}{z \times (log z)^2}\times \frac{dz}{dx}+\frac{1}{x logz}=\frac{1}{x^2}$→→Dividing both sides by z. (log z)²,we have got this expression

→Put , $\frac{1}{logz}= t$

Differentiating both sides

$\frac{dt}{dz}=\frac{-1}{z.(log z)^2} \\\\ \frac{dz}{dx}\times \frac{-1}{z.(log z)^2}=\frac{dt}{dx}$

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

Now, this is linear differential equation

Integrating Factor = $e^{\int \frac{1}{x} \, dx =e^{log x}=x$

Multiplying both sides by x, we get

$x \times[\frac{dt}{dx}+ \frac{t}{x}]=\frac{1}{x}$

Now, integrating both sides

x t= log x + K, where K is any constant because , $\int\limits\frac{1} {x} \, dx =log x + K$