Question

Let $y\left(x\right)$ is the solution of the differential equation $\left(x + 2\right)\frac{d y}{d x}-\left(x + 1\right)y=2$ . If $y\left(0\right)=-1$ , then the value of $y\left(2\right)$ is equal to

• A. $e^{2}+\frac{1}{2}$
• B. $e^{2}-\frac{1}{2}$
• C. $\frac{1}{2}-e^{2}$
• D. $e^{2}$

Answer 1

Answer: (B)

Given differential equation is $\frac{d y}{d x}-\frac{x + 1}{x + 2}y=\frac{2}{x + 2}$
I.F. $=e^{- \displaystyle \int \frac{x + 1}{x + 2} d x}=e^{- x}\left(x + 2\right)$
So the solution of the equation is $y\left(x + 2\right)e^{- x}=-2e^{- x}+C$
Since $y\left(0\right)=1,$ we have $C=4$
So $y\left(2\right)=e^{2}-\frac{1}{2}$