Question

solve the given differential equation.
\\(\frac{dy}{dx}\frac{ln(x)}{y}=(\frac{y + 1}{x})^2\\)

Explanation:

Step1: Rearrange the equation

Separate variables: $\frac{y\ln(x)}{y + 1}dy=\frac{x}{y}dx$. Cross - multiply to get $y^{2}\ln(x)dy=x(y + 1)dx$. Then $\frac{y^{2}}{y + 1}dy=\frac{x}{\ln(x)}dx$.

Step2: Simplify the left - hand side

Perform polynomial long division on $\frac{y^{2}}{y + 1}$. $y^{2}=(y + 1)(y-1)+1$, so $\frac{y^{2}}{y + 1}=y - 1+\frac{1}{y + 1}$.

Step3: Integrate both sides

Integrate $\int(y - 1+\frac{1}{y + 1})dy=\int\frac{x}{\ln(x)}dx$.
For the left - hand side: $\int(y - 1+\frac{1}{y + 1})dy=\frac{y^{2}}{2}-y+\ln|y + 1|+C_1$.
For the right - hand side, let $u = \ln(x)$, then $du=\frac{1}{x}dx$ and $x = e^{u}$. So $\int\frac{x}{\ln(x)}dx=\int\frac{e^{u}}{u}\cdot e^{u}du=\int\frac{e^{2u}}{u}du$. But a more common way is to use substitution for $\int\frac{x}{\ln(x)}dx$. Let $t=\ln(x)$, then $x = e^{t}$ and $dx=e^{t}dt$. So $\int\frac{x}{\ln(x)}dx=\int\frac{e^{t}}{t}\cdot e^{t}dt$. Another way is $\int\frac{x}{\ln(x)}dx=\int\frac{1}{\ln(x)}d(\frac{x^{2}}{2})$. Using integration by parts $\int\frac{x}{\ln(x)}dx=\frac{x^{2}}{2\ln(x)}+\frac{1}{2}\int\frac{x}{\ln(x)}dx-\frac{1}{2}\int\frac{x}{(\ln(x))^{2}}dx$. A better substitution for $\int\frac{x}{\ln(x)}dx$ is to let $u=\ln(x)$, $x = e^{u}$, $dx=e^{u}du$, then $\int\frac{x}{\ln(x)}dx=\int\frac{e^{u}}{u}\cdot e^{u}du$. However, if we rewrite the original separation as $\frac{y(y + 1)}{x}dy=\frac{1}{\ln(x)}dx$ and integrate:
$\int\frac{y^{2}+y}{x}dy=\int\frac{1}{\ln(x)}dx$. Integrating $\int(y^{2}+y)dy=\int\frac{x}{\ln(x)}dx$. $\int(y^{2}+y)dy=\frac{y^{3}}{3}+\frac{y^{2}}{2}+C_1$.
For $\int\frac{x}{\ln(x)}dx$, let $u = \ln(x)$, $x=e^{u}$, $dx = e^{u}du$, then $\int\frac{x}{\ln(x)}dx=\int\frac{e^{u}}{u}\cdot e^{u}du$. A more straightforward substitution for $\int\frac{x}{\ln(x)}dx$ is to use $u=\ln(x)$, $du=\frac{1}{x}dx$, and $\int\frac{x}{\ln(x)}dx=\int\frac{1}{u}du=\ln|\ln(x)|+C_2$.
So $\frac{y^{2}}{2}-y+\ln|y + 1|=\ln|\ln(x)|+C$.

Answer:

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