CUET Preparation Today
Integrating factor of the differential equation $ (x+1) \frac{dy}{dx}-y = e^{3x}(x=1)^2,$ is |
$-(x+1)$ $log(x+1)$ $e^{x+1}$ $\frac{1}{x+1}$ |
$\frac{1}{x+1}$ |
The correct answer is option (4) : $\frac{1}{x+1}$ $\frac{dy}{dx} -\frac{1}{x+1}y=e^{3x}(x+1)$ $∴I.F. = e^{-∫\frac{1}{x+1}dx}=e^{-log(x+1)}=\frac{1}{x+1}$ |
