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Question
Find the equation of a curve passing through the point (0, 0) and whose differential equation is y′ = e x sin x.
Solution
We have,
y' = ex 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 = -ex cos x + ex sin x + 1
Hence, 2y - 1 = ex (sin x - cos x)
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