Question

Determine an appropriate substitution to solve

$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathit{y}\mathit{I}\mathit{n}\left(xy\right)$

Short answer

The equation is$\mathit{y}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}\mathbf{c}\mathbf{-}\mathbf{1}}}{\mathbf{x}}$

Step-by-step solution

The differential equation

The given differential equation, can be written as

$y\text{'}=\frac{y}{x}In\left(xy\right)$

Let $u=xy\Rightarrow y=\frac{u}{x}$, then

$y\text{'}=\frac{u\text{'}x-u}{x^2}$

Substitute$y=\frac{u}{x}$

$$\begin{gathered} \frac{u\text{'}x-u}{x^2}=\frac{{\displaystyle \frac{u}{x}}}{x}Inu \\ \frac{u\text{'}x-u}{x^2}=\frac{{\displaystyle u}}{x^2}Inu \end{gathered}$$

Multiply by x²

$$\begin{gathered} u\text{'}x-u=uInu \\ u\text{'}x=uInu+u \\ u\text{'}x=u\left(Inu+1\right) \\ \frac{du}{dx}x=u\left(Inu+1\right) \\ \frac{du}{u\left(Inu+1\right)}=\frac{dx}{x} \end{gathered}$$

Separate variables.

Each Side Integrate

$$\begin{gathered} \frac{du}{u\left(Inu+1\right)}=\frac{dx}{x} \\ Inu=z\Rightarrow \frac{du}{u}=dz \\ \frac{dz}{z+1}=In\left|x\right|+In\left|c\right| \\ In\left|z+1\right|=In\left|xc\right|\left(Inab=Ina+Inb\right) \\ z+1=xc \\ z=xc-1 \\ Inu=xc-1\left(u=xy\right) \\ In\left(xy\right)=xc-1 \\ xy=e^{xc-1} \\ y=\frac{e^{xc-1}}{x} \end{gathered}$$