Q.Solve the following differential equation: $\frac{dy}{dx} + 2y = \sin x$

The key idea is solving a first-order linear ODE using an integrating factor.

Step 1 – Identify the integrating factor.

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = 2$.

Integrating factor:

Step 2 – Multiply through and integrate.

Multiplying the equation by $e^{2x}$:

Integrate both sides:

Step 3 – Evaluate the integral.

Using integration by parts (or the standard formula):

Step 4 – Solve for $y$.

Divide through by $e^{2x}$:

Why This Approach Works

The equation $\frac{dy}{dx} + 2y = \sin x$ is a first-order linear ordinary differential equation — it has the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Here $P(x) = 2$ (constant) and $Q(x) = \sin x$.

The key insight: the left side $\frac{dy}{dx} + 2y$ looks almost like the derivative of a product. If we multiply both sides by a cleverly chosen function $\mu(x)$, the left side becomes exactly $\frac{d}{dx}(\mu y)$. That function is the integrating factor.

Once we multiply through by $\mu$, we can integrate both sides directly — no guesswork needed.

Step-by-Step Solution

1. Identify $P(x)$ and compute the integrating factor.

Here $P(x) = 2$, so:

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

Notice the left side is now exactly $\frac{d}{dx}\left(e^{2x}y\right)$ — check by differentiating: $\frac{d}{dx}(e^{2x}y) = e^{2x}\frac{dy}{dx} + 2e^{2x}y$. Perfect.

3. Rewrite and integrate.

Integrate both sides with respect to $x$:

4. Evaluate the integral $\int e^{2x}\sin x\,dx$.

This is a classic integration by parts (or use the formula for $\int e^{ax}\sin(bx)\,dx$). Let's do it cleanly.

So:

5. Solve for $y$.

From step 3:

Divide through by $e^{2x}$ (which is never zero):

Method: Integrating factor for a linear equation with a trig right-hand side

Use the integrating-factor method for any first-order equation of the shape $\dfrac{dy}{dx}+P\,y=Q$, where $y$ and $\dfrac{dy}{dx}$ appear only to the first power.

Steps

Step 1: Put it in standard form.

Make the coefficient of $\dfrac{dy}{dx}$ equal to $1$, then read off $P$ and $Q$.

Step 2: Build the integrating factor.

Multiplying by it makes the left side an exact derivative $\dfrac{d}{dx}\big(y\cdot\text{I.F.}\big)$.

Step 3: Integrate $Q\times\text{I.F.}$.

Step 4: Handle an $e^{ax}\sin bx$ or $e^{ax}\cos bx$ integral.

Such integrals return to themselves after integrating by parts twice; solve the resulting algebra for the integral, or use $\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}$. Divide out the I.F. to finish.

Common Mistakes

Mistake 1: Stopping at $y\,e^{2x}=\int e^{2x}\sin x\,dx$ without the constant.

Why it's wrong: the arbitrary constant $C$ must appear at the integration step; omitting it loses the general solution. Correct approach: write $y\,e^{2x}=\dfrac{e^{2x}(2\sin x-\cos x)}{5}+C$.

Mistake 2: Sign errors in $\int e^{2x}\sin x\,dx$.

Why it's wrong: this integral returns to itself after two by-parts steps; a flipped sign gives $2\sin x+\cos x$ instead of $2\sin x-\cos x$. Correct approach: use $\int e^{ax}\sin bx\,dx=\dfrac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}$ with $a=2,b=1$.

Mistake 3: Forgetting the $Ce^{-2x}$ term after dividing by the I.F.

Why it's wrong: dividing by $e^{2x}$ turns $C$ into $Ce^{-2x}$, a genuine part of the solution. Correct approach: divide every term, including the constant.