Question

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

Answer 1

We have,

y' = e^x sin x

⇒ `dy/dx = e^x sin x`

⇒ `dy = e^x sin x dx`                    ....(1)

Integrating (1) both sides, we get

`int dy = inte^x sin x dx`

⇒ `y = -e^x cos x + int e^x cos x dx`

⇒ `y = -e^x cos x + e^x sin x - int e^x sin x dx`

⇒ `y = - e^x cos x +e^x sin x - y + C`

⇒ `2y = -e^x cos x + e^x sin x + C`

As pint (0, 0) lies on it, i.e., x = 0, y = 0

∴ 0 = -e⁰ + C

⇒ C = 1

∴ Required equation is

2y = -e^x  cos x + e^x sin x + 1

Hence, 2y - 1 = e^x (sin x - cos x)