# Solve the following differential equation : $\large\frac{dy}{dx}$$+y\cot\: x = x^2\cot x+2x ## 1 Answer Toolbox: • To solve the first order linear differential equation of the form \large\frac{dy}{dx}$$ + Py = Q$
• (i) Write the given equation in the form of $\large\frac{dy}{dx}$$+ Py = Q • (ii) Find the integrating factor (I.F) = e^{\int Pdx}. • (iii) Write the solution as y(I.F) = integration of Q(I.F) dx + C Step 1: \large\frac{dy}{dx}$$+y\cot x=2x+x^2\cot x$
Clearly the given equation is of the form
$\large\frac{dy}{dx}$$+Py=Q The solution for the above equation is y.e^{\int Pdx}=\int Q.e^{\int Pdx}dx+c Here P=\cot x Q=2x+x^2\cot x \int Pdx=\log \sin x e^{\large\int Pdx}=e^{\large\log \sin x}=\sin x Step 2: y\sin x=\int (2x+x^2\cot x).\sin xdx+c \qquad\;\;=\int 2x\sin xdx+\int x^2\cot x.\sin xdx+c \qquad\;\;=\int 2x\sin xdx+\int x^2\cos x.dx+c Step 3: Consider I=2\int x\sin xdx This is of the form \int u dv=uv-\int vdu Let du=x u=\large\frac{x^2}{2}$$dx$
$v=\sin x$
$dv=\cos xdx$
$I=-\sin x\big(\large\frac{2x^2}{2}\big)$$-\int\cos x\big(\large\frac{2x^2}{2}\big)$$dx+\int x^2\cos xdx+c$
$y\sin x=x^2\sin x-\int x^2\cos xdx+\int x^2\cos xdx+c$
$y\sin x=x^2\sin x+c$
This is the required solution.