Q.(ix) General solution of $\frac{dy}{dx}+y=\sin x$ is ______.

The key idea is the Integrating Factor Method for a first-order linear ODE.

Step 1: The equation is $\frac{dy}{dx} + y = \sin x$. Here $P(x) = 1$ and $Q(x) = \sin x$.

Step 2: Compute the integrating factor:

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

Step 3: Multiply through and integrate:

$e^x \frac{dy}{dx} + e^x y = e^x \sin x$

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

$y e^x = \int e^x \sin x \, dx$. …

The equation $\frac{dy}{dx} + y = \sin x$ is a classic first-order linear ordinary differential equation. The standard form is $\frac{dy}{dx} + P(x) y = Q(x)$, where here $P(x) = 1$ and $Q(x) = \sin x$.

Why does the Integrating Factor method work? The left side $\frac{dy}{dx} + y$ is almost the derivative of a product — but not quite. If we multiply the whole equation by a cleverly chosen function $\mu(x)$, the left side becomes exactly $\frac{d}{dx}(\mu y)$. That function is the integrating factor, given by $\mu(x) = e^{\int P(x) dx}$.

1. Find the integrating factor. Since $P(x) = 1$, we have $\int P(x) dx = \int 1 dx = x$. So the integrating factor is:

2. Multiply the ODE by $\mu(x)$. Multiplying both sides of $\frac{dy}{dx} + y = \sin x$ by $e^x$ gives:

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

3. Rewrite and integrate. The equation becomes:

Integrate both sides with respect to $x$:

where $C$ is the constant of integration.

4. Evaluate the integral $\int e^x \sin x \, dx$. This is a standard cyclic integral. Use integration by parts twice, or recall the formula:

Here $a = 1$, $b = 1$, so:

$\int e^x \sin x \, dx = \frac{e^x}{1^2 + 1^2} (1 \cdot \sin x - 1 \cdot \cos x) = \frac{e^x}{2} (\sin x - \cos x).$ …

Method: Integrating Factor with a Constant Coefficient

For a linear equation $\frac{dy}{dx}+ky=Q(x)$ with $P$ constant, the integrating factor is simply $e^{kx}$, and the work shifts to integrating $Q(x)\,e^{kx}$.

Steps

Step 1: Read off $P$ and $Q$.

With $\frac{dy}{dx}+P\,y=Q(x)$ and $P$ a constant, the integrating factor is

Step 2: Reduce to a single derivative and integrate.

Step 3: Evaluate the exponential–trig integral by the standard result. …

Common Mistakes

Mistake 1: Writing the integral of $e^x\sin x$ as $e^x\cos x$ or $-e^x\cos x$.

Why it's wrong: $e^x\sin x$ is a cyclic integral requiring integration by parts twice; the correct result is $\frac{e^x}{2}(\sin x-\cos x)$. Correct approach: use $\int e^{ax}\sin bx\,dx=\frac{e^{ax}}{a^2+b^2}(a\sin bx-b\cos bx)$ with $a=b=1$.

Mistake 2: Dropping the $Ce^{-x}$ term after dividing by $e^x$. …