Question

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

• A. $xy = - e^x$
• B. $xy = - e^{-x}$
• C. $xy = - 1$
• D. $y = 2 \,e^x - 1$

Answer 1

Answer: (D)

$y = 2 \,e^x - 1$

Answer 2

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