Q.Find the general solution of $\frac{dy}{dx}+ay=e^{mx}$.

We are solving a first-order linear differential equation — an Initial Value Problem once a condition is given, but here we only need the general solution.

Step 1: Identify the integrating factor.

The standard form is $\frac{dy}{dx} + P(x)y = Q(x)$. Here $P(x) = a$ (constant), so the integrating factor is

Step 2: Multiply through and integrate.

Integrate both sides:

Step 3: Evaluate the integral — two cases.

• If $a+m \neq 0$: …

The equation $\frac{dy}{dx} + a y = e^{mx}$ is a classic first-order linear ordinary differential equation. The structure is: derivative of $y$ plus a constant times $y$ equals an exponential. The key idea is to multiply both sides by an integrating factor — a function that turns the left-hand side into the derivative of a product. This works because the left side already looks like the result of a product rule if we choose the right factor.

Why does this work? For any linear ODE of the form $\frac{dy}{dx} + P(x) y = Q(x)$, the integrating factor is $\mu(x) = e^{\int P(x) dx}$. Here $P(x) = a$ (constant), so $\mu(x) = e^{\int a \, dx} = e^{ax}$. Multiplying through gives $\frac{d}{dx}(y e^{ax}) = e^{(a+m)x}$, which is straightforward to integrate.

Let’s go step by step.

1. Identify the standard form and find the integrating factor.

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

   The integrating factor is $\mu(x) = e^{\int a \, dx} = e^{ax}$.

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

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

4. Integrate both sides with respect to $x$.

The integral depends on whether $a+m = 0$ or not — this is the critical branching point.

• Case 1: $a + m \neq 0$

 So $y e^{ax} = \frac{e^{(a+m)x}}{a+m} + C$.

• Case 2: $a + m = 0$ (i.e., $m = -a$) Then $e^{(a+m)x} = e^{0} = 1$, so

 So $y e^{ax} = x + C$.

5. Solve for $y$ in each case.

• For $a+m \neq 0$: …

Method: Linear DE with constant coefficient and exponential forcing

Use this for $\dfrac{dy}{dx}+ay=e^{mx}$ and any linear first-order equation with a constant coefficient.

Steps

Step 1: Identify $P$ and the integrating factor

Here $P=a$ is constant, so

Step 2: Multiply through and recognise the exact derivative

$\frac{d}{dx}\big(y\,e^{ax}\big)=e^{ax}\cdot e^{mx}=e^{(a+m)x}.$ …

Common Mistakes

Mistake 1: Using $e^{\int a\,dx}=e^{a}$ instead of $e^{ax}$

Why it's wrong: $a$ is constant, so $\int a\,dx=ax$, giving $\mathrm{IF}=e^{ax}$. Correct approach: integrate $P$ with respect to $x$.

Mistake 2: Ignoring the special case $a+m=0$ …