Question

What is the solution of the DE $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{yf}\text{'}\left(x\right)-y^2}{f\left(x\right)}?$

• A. $f\left(x\right)+y=c$
• B. $f\left(x\right)=y(x+c)$
• C. $y=\frac{c}{f\left(x\right)}+1$
• D. $y=\mathrm{xf}\left(x\right)+c$

Answer 1

The correct option is B $f\left(x\right)=y(x+c)$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{yf}\text{'}\left(x\right)-y^2}{f\left(x\right)}\left(\mathrm{yf}\text{'}\left(x\right)-y^2\right)\mathrm{dx}=f\left(x\right)\mathrm{dy}\frac{\mathrm{yf}\text{'}\left(x\right)\mathrm{dx}-f\left(x\right)\mathrm{dy}}{y^2}=\mathrm{dx}\mathrm{This}\mathrm{is}\mathrm{an}\mathrm{exact}\mathrm{form}d\left(\frac{f\left(x\right)}{y}\right]=d\left(x\right)\mathrm{Integrating}\frac{f\left(x\right)}{y}=x+cf\left(x\right)=y\left(x+c\right)$