Question

Consider the differential equation $\frac{dy}{dx}+y=e^x$ with y(0) = 1. Then the value of y(1) is

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

Answer 1

The correct option is C $\frac{1}{2}[e+e^{-1}]$

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

It is LDE in y & x

I.F = $e^{\int dx}=e^x$

$\therefore$ The general solution of equation (1) is

y(I.F) = $\int (I.F)e^{x}dx+C$

$y.e^x=\int e^x{e}^{x}dx+C$

y.$e^x=\frac{e^{2x}}{2}+C...\left(2\right)$

Now using y(0) = 1 $\Rightarrow C=\frac{1}{2}$

Hence Solution is

$y{e}^x=\frac{e^{2x}}{2}+\frac{1}{2}$

$\therefore y\left(1\right)=\frac{e}{2}+\frac{e^{-1}}{2}=\frac{e+e^{-1}}{2}$