Q.The solution of $\frac{dy}{dx}+y=e^{-x}$, $y(0)=0$ is:
(A) $y=e^x(x-1)$
(B) $y=xe^{-x}$
(C) $y=xe^{-x}+1$
(D) $y=(x+1)e^{-x}$

This is an Initial Value Problem — a first-order linear ODE with a given condition.

Step 1: Identify integrating factor.

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$. Here $P(x)=1$, so integrating factor $\mu = e^{\int 1\,dx} = e^x$.

Step 2: Multiply through and integrate.

$e^x \frac{dy}{dx} + e^x y = e^x \cdot e^{-x} = 1$

$\Rightarrow \frac{d}{dx}(y e^x) = 1$

Integrate: $y e^x = x + C$. …

The key idea: when you see $\frac{dy}{dx} + P(x)y = Q(x)$, the standard tool is the integrating factor. It rewrites the left side as a perfect derivative, turning the problem into a direct integration.

Here, $P(x) = 1$ and $Q(x) = e^{-x}$. The integrating factor is $e^{\int 1\,dx} = e^x$. Multiply through, and the left side becomes $\frac{d}{dx}(y e^x)$. Then integrate both sides, apply the initial condition, and solve for $y$.

Let’s walk through it step by step.

1. Identify the form and find the integrating factor

   The equation is $\frac{dy}{dx} + y = e^{-x}$. This is linear with $P(x)=1$.

   The integrating factor $\mu(x) = e^{\int P\,dx} = e^{\int 1\,dx} = e^x$.

2. Multiply the entire equation by $\mu(x)$

The right side simplifies: $e^x \cdot e^{-x} = e^0 = 1$.

So we have:

3. Recognize the left side as a derivative Notice that $\frac{d}{dx}(y e^x) = e^x \frac{dy}{dx} + e^x y$. So the equation becomes:

4. Integrate both sides

This gives:

where $C$ is the constant of integration.

5. Apply the initial condition $y(0)=0$ …

Method: Solving a Linear IVP by Integrating Factor

For a first-order linear equation with an initial condition, build the integrating factor, integrate to the general solution, then fix the constant from the condition.

Steps

Step 1: Form the integrating factor.

For $\frac{dy}{dx}+y=e^{-x}$, $P=1$ so $\text{I.F.}=e^{x}$.

Step 2: Reduce to a derivative and integrate.

$\frac{d}{dx}(y\,e^{x})=e^{x}\cdot e^{-x}=1\ \Rightarrow\ y\,e^{x}=x+C.$ …