Q.Find the equation of a curve passing through the point $(0, 0)$ and whose differential equation is $y' = e^x \sin x$.

Concept: Separation of Variables — the equation is already in separated form $dy = e^x \sin x \, dx$, so we integrate directly.

Step 1: Write the differential equation as

Step 2: Integrate both sides:

Use integration by parts (or the standard formula $\int e^{ax}\sin bx\,dx = \frac{e^{ax}(a\sin bx - b\cos bx)}{a^2+b^2}$). Here $a=1$, $b=1$:

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

The differential equation given is $y' = e^x \sin x$. This is a first-order ODE where the derivative is expressed purely in terms of $x$ — there is no $y$ on the right-hand side. That makes it a separable equation in the simplest sense: we can directly integrate.

The core idea of Separation of Variables is to rearrange the equation so that all terms involving $y$ are on one side and all terms involving $x$ are on the other. Here, since $y' = \frac{dy}{dx}$, we write:

Multiply both sides by $dx$:

Now the variables are separated — $y$ on the left, $x$ on the right. The next step is to integrate both sides.

1. Integrate both sides

The left side is simply $y + C_1$. The right side requires integration by parts (or a standard formula).

2. Evaluate $\int e^x \sin x \, dx$

   Let $I = \int e^x \sin x \, dx$. Use integration by parts twice.

   First, let $u = \sin x$, $dv = e^x dx$. Then $du = \cos x \, dx$, $v = e^x$.

Now evaluate $\int e^x \cos x \, dx$. Again, let $u = \cos x$, $dv = e^x dx$, so $du = -\sin x \, dx$, $v = e^x$.

But $\int e^x \sin x \, dx$ is exactly $I$. So we have:

Simplify:

Bring $I$ terms together:

Therefore:

$I = \frac{e^x}{2} (\sin x - \cos x) + C$ …

Method: Direct integration of a product $e^x\sin x$ via by-parts recursion

Use this when the curve's slope is given as $y'=f(x)$ and $f(x)$ is a product like $e^x\sin x$, whose integral returns to itself under repeated by-parts.

Steps

Step 1: Set up direct integration

Since $y'=e^x\sin x$, write $y=\int e^x\sin x\,dx$.

Step 2: Integrate by parts twice

Apply integration by parts twice; the original integral $I$ reappears, letting you solve algebraically:

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

Common Mistakes

Mistake 1: Abandoning the by-parts recursion too early

Why it's wrong: $\int e^x\sin x\,dx$ reappears after two by-parts steps; stopping after one leaves an unfinished integral. Correct approach: apply by-parts twice, then solve algebraically for the integral.

Mistake 2: Sign errors between $\sin x$ and $\cos x$ terms …