Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. Solution of $\frac{dy}{dx}-y=1$, $y\left(0\right)=1$ is given by

Differential Equations

Solution:

$\frac{dy}{dx}-y=1$
$\Rightarrow \frac{dy}{dx}=1+y$
$\Rightarrow \frac{dy}{1+y}=dx$
On integrating, we get $log\left( 1 + y\right) = x + c$
Now, $y\left(0\right)=1$
$\Rightarrow log2=c$
$\therefore log\left(1 + y\right) = x + log2$
$\Rightarrow \frac{1+y}{2}=e^{x}$
$\Rightarrow 1+y=2e^{x}$
$\Rightarrow y=2e^{x}-1$