Question

Solution of the differential equation $(1+x^2)dy+2xydx=\cot xdx$ is

• A. $y=\log |\sin x|(1+x^2)+C(1+x^2{)}^{-1}$
• B. $y=\log |\sin x|(1+x^2{)}^{-1}+C(1+x^2)$
• C. $y=\log |\sin x|(1+x^2)+C(1+x)$
• D. $y=\log |\sin x|(1+x^2{)}^{-1}+C(1+x^2{)}^{-1}$

Answer 1

The correct option is D $y=\log |\sin x|(1+x^2{)}^{-1}+C(1+x^2{)}^{-1}$

$(1+x^2)dy+2xydx=\cot xdx$

Put in form $\frac{dy}{dx}+PY=Q$

$(1+x{)}^{2}dy+2xydx=\cot xdx$

dividing both sides by $(1+x^2)$

$\frac{dy}{dx}+\frac{2xy}{1x^2}\frac{dx}{dx}=\frac{\cot x}{(1+x^2)}\frac{dx}{dx}$

$\frac{dy}{dx}+\left(\frac{2x}{1+x^2}\right)y=\frac{\cot x}{1+x^2}$

Find P and Q

$\frac{dy}{dx}+PY=Q$

$P=\frac{2x}{1+x^2}$

$Q=\frac{\cot x}{1+x^2}$

If$=e^{\int \frac{2x}{1+x^2}}$ $t=1+x^2$

If $=e^{\int \frac{dt}{t}}=e^{logt}=t=1+x^2$

Solution: $Y\times IF=\int Q\times IFdx+C$

$Y(1+x^2)=\int \frac{\cot x}{1+x^2}\times (1+x^2)dx+c$

$Y(1+x^2)=\int \cot xdx+c$

$Y(1+x^2)=log|\sin x|+c$

$Y=(1+x^2{)}^{-1}log|\sin x|+c(1+x^2{)}^{-1}$.