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# y’ = x + 4y cot 2x (using linear DE)?

### 1 Answer

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- Ian HLv 74 weeks ago
Here is a start; you join in.

dy/dx = x + 4y cot(2x)

dy/dx – [4/tan(2x)]y = x

HINT: Integrating factor I = e^∫– [4/tan(2x)]dx

∫– [4/tan(2x)]dx = -2ln[sin(2x)] = ln{1/[sin(2x)]^2}

I = e^∫P(x)dx = 1/[sin(2x)]^2 or [cosec(2x)]^2 if you prefer.

Form an exact equation on the left and integrate

Use I(x)*y = ∫x*I(x) dx

See the next step and view the solution you should get at

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