The solution of (dy/dx)+y=e-x, y(0)=0 is

Question

The solution of (dy/dx)+y=e-x, y(0)=0 is (A) y = e-x (x - 1) (B) y = xex (C) y = xe-x + 1 (D) y=xe-x

Tardigrade Admin · Accepted Answer

(dy/dx)+y=e-x It is linear differential equation with I.F.=e∫ dx=ex Now, solution is y⋅ ex=∫ ex e-x dx+c ⇒ yex = x + c ⇒ y = (x + c)e-x Now, y(0)=0 ⇒ c=0 Hence, y = xe-x

Q. The solution of $\frac{d y}{d x} + y = e^{- x}$ , $y (0) = 0$ is

$y = e^{- x} (x - 1)$

$y = x e^{x}$

$y = x e^{- x} + 1$

$y = x e^{- x}$

$\frac{d y}{d x} + y = e^{- x}$
It is linear differential equation with
$I . F . = e^{\int d x} = e^{x}$
Now, solution is $y \cdot e^{x} = \int e^{x} e^{- x} d x + c$
$\Rightarrow y e^{x} = x + c$
$\Rightarrow y = (x + c) e^{- x}$
Now, $y (0) = 0$
$\Rightarrow c = 0$
Hence, $y = x e^{- x}$

Q. The solution of dxdy​+y=e−x, y(0)=0 is

y=e−x(x−1)

y=xex

y=xe−x+1

y=xe−x