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) $y=xe^{-x}$

This is a first-order linear differential equation — an Initial Value Problem.

Step 1 – Identify integrating factor

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

Integrating factor: $\mu = e^{\int 1\,dx} = e^{x}$.

Step 2 – Multiply and integrate

Multiplying: $e^{x}\frac{dy}{dx} + e^{x}y = 1$

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

Integrate: $y e^{x} = x + C$. …

The equation $\frac{dy}{dx} + y = e^{-x}$ with $y(0)=0$ is a classic Initial Value Problem (IVP). The key idea: when you have a first-order linear ODE of the form $\frac{dy}{dx} + P(x)y = Q(x)$, you can multiply both sides by an integrating factor — a function that turns the left side into the derivative of a product. This makes the equation directly integrable.

Why does this work? Because the left side $\frac{dy}{dx} + y$ looks almost like the derivative of $y$ times something, but it's missing the derivative of that "something". The integrating factor supplies exactly that missing piece.

Let's walk through it.

1. Identify the standard form.

   The equation is already in the form $\frac{dy}{dx} + P(x)y = Q(x)$, with $P(x) = 1$ and $Q(x) = e^{-x}$.

2. Compute the integrating factor.

   The integrating factor $\mu(x)$ is given by $e^{\int P(x)\,dx}$.

   Here $\int 1\,dx = x$, so

3. Multiply the entire ODE by $\mu(x)$.

Notice the left side is now exactly $\frac{d}{dx}(e^{x} y)$, because by the product rule:

4. Rewrite and integrate.

Integrate both sides with respect to $x$:

where $C$ is the constant of integration.

5. Solve for $y$. …

Method: Solve a linear first-order IVP, then apply the initial condition

Use this whenever a linear equation $\frac{dy}{dx}+P(x)y=Q(x)$ comes with a condition like $y(0)=0$.

Steps

Step 1: Find the general solution by the integrating factor.

With $P=1$, the I.F. is $e^{x}$, the left side becomes $\frac{d}{dx}(ye^{x})$, and integrating gives $ye^{x}=\int Q\,e^{x}\,dx+C$.

Step 2: Keep the constant $C$ — do not drop it.

The general solution must still contain $C$ at this stage; the initial condition is what pins it down. …

Common Mistakes

Mistake 1: Forgetting to apply the initial condition.

Why it's wrong: stopping at $y=e^{-x}(x+C)$ leaves the answer with an undetermined constant (matching option A). Correct approach: substitute $x=0,\ y=0$ to find $C=0$, giving $y=xe^{-x}$.

Mistake 2: Using the wrong integrating factor. …