Question

In Problems 1–22 Solve the given differential equation by separation of variables.

$\mathbf{\text{ylnx}}\frac{\mathbf{\text{dx}}}{\mathbf{\text{dy}}}\mathbf{\text{=}}{\left(\frac{{\displaystyle \text{y+1}}}{{\displaystyle \text{x}}}\right)}^{\mathbf{\text{2}}}$

Short answer

$\frac{1}{3}{x}^3\ln x-\frac{1}{9}{x}^3=\frac{y^2}{2}+2y+\ln \left|y\right|+C$

Step-by-step solution

Step 1:Definition of separable equation

A first-order differential equation of the form $\frac{dy}{dx}=g\left(x\right)h\left(y\right)$is said to be separable or to have separable variables.

Step 2:Separate the variables

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

Rearrange

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

Separate the variables

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

Expand${(y+1)}^2$

$x^2\ln xdx=\frac{y^2+2y+1}{y}dy$

Step 3:Use integration by parts

Integrate both sides

$\int x^2\ln xdx=\int \frac{y^2+2y+1}{y}dy$

Integrating by parts

$$\begin{gathered} \int x^2\ln xdx \\ u=\ln x\Rightarrow du=\frac{dx}{x} \\ dv=x^{2}dx\Rightarrow v=\frac{x^3}{3} \end{gathered}$$

So

$\underset{\int udv}{\underbrace{\int x^2\ln xdx}}=\underset{uv}{\underbrace{\frac{x^3\ln x}{3}}}-\underset{\int vdu}{\underbrace{\int \left(\frac{x^3}{3}\right)\left(\frac{dx}{x}x\right)}}$

$$\begin{gathered} =\frac{x^3\ln x}{3}-\frac{1}{3}\int x^{2}dx \\ =\frac{x^3\ln x}{3}-\frac{1}{3}\left(\frac{{\displaystyle x^3}}{{\displaystyle 3}}\right)+C \\ =\frac{1}{3}{x}^3\ln x-\frac{1}{9}{x}^3+C \end{gathered}$$

Step 3:Derive integration

Integrating

$\int \frac{y^2+2y+1}{y}dy$

$\frac{y^2+2y+1}{y}dy=\int (\frac{y^2}{y}+\frac{2y}{y}+\frac{1}{y})dy$

$=\int (y+2+\frac{1}{y})dy$

$=\frac{1}{2}{y}^2+2y+\ln \left|y\right|+C$

Therefore the solution is$\frac{\text{1}}{\text{3}}{\text{x}}^{\text{3}}\text{lnx-}\frac{\text{1}}{\text{9}}{\text{x}}^{\text{3}}\text{=}\frac{{\text{y}}^{\text{2}}}{\text{2}}\text{+2y+ln|y|+C}$