Solve the following differential equation: dy/dx + y cot x = x^2 cot x + 2x

Question

Solve the following differential equation:

\(\cfrac{dy}{dx}\) + y cot x = x² cot x + 2x

Answer 1

 \(\cfrac{dy}{dx}\) + y cot x = x² cot x + 2x ........ (1)

This is the linear differential equation of the form

\(\cfrac{dy}{dx}\) + Py = Q, where P = cot x and Q = x² cot x + 2x

∴ I.F. = e^{∫ Pdx}

= e^{∫ cot xdx}

= e^{log(sin x)}

= sin x

∴ the solution of (1) is given by

y(I.F.) = ∫Q . (I.F.) dx + c

∴ y sin x = ∫(x² cot x + 2x) sin x dx + c

∴ y sin x = ∫(x² cot x . sin x + 2x sin x) dx + c

∴ y sin x = ∫x² cos x dx + 2∫x sin x dx + c

∴ y sin x = x² ∫cos x dx – ∫[\(\cfrac{d}{dx}\)(x²) ∫cos x dx] dx + 2∫x sin x dx + c

∴ y sin x = x² (sin x) – ∫2x(sin x) dx + 2∫x sin x dx + c

∴ y sin x = x² sin x – 2∫x sin x dx + 2∫x sin x dx + c

∴ y sin x = x² sin x + c

∴ y = x² + c cosec x

This is the general solution.