Question

The solution of $\frac{dy}{dx}+x=x{e}^{(n-1)y}$ is:

• A. $\frac{1}{n-1}\log \left(\frac{e^{(n-1)y}-1}{e^{(n-1)y}}\right)=x^2/2+C$
• B. $e^{(n-1)y}=C{e}^{(n-1)y+(n-1)x^{2/2}}+1$
• C. $\log \left(\frac{e^{(n-1)y}-1}{(n-1)e^{(n-1)y}}\right)=x^2+C$
• D. $e^{(n-1)}y=c{e}^{(n-1)y^{2/2+x}}+1$

Answer 1

Answer: (A), (B)

The correct options are
A $\frac{1}{n-1}\log \left(\frac{e^{(n-1)y}-1}{e^{(n-1)y}}\right)=x^2/2+C$
B $e^{(n-1)y}=C{e}^{(n-1)y+(n-1)x^{2/2}}+1$

Given $\frac{dy}{dx}+x=x{e}^{y(n-1)}$

$\Rightarrow \frac{dy}{dx}=x(e^{y(n-1)}-1)$

$\Rightarrow \frac{dy}{(e^{y(n-1)}-1)}=xdx$

integrating on both sides

$\Rightarrow \frac{1}{n-1}\int \frac{(n-1)}{(e^{y(n-1)}-1)e^{y(n-1)}}{e}^{y(n-1)}dy=\frac{x^2}{2}+c$

$\Rightarrow e^{y(n-1)}=c{e}^{(n-1)+(n-1)\frac{x^2}{2}}+1$