Question

The solution of the differential equation $\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{si{n}^{2}x}{1+x^3}$ is

• A. $y(1+x^3)=x+\frac{1}{2}sin2x+c$
• B. $2ta{n}^{-1}\left(\frac{x}{y}\right)+logx+c=0$
• C. $log(y+\sqrt{x^2+y^2})+logy+c=0$
• D. $sin{h}^{-1}\left(\frac{x}{y}\right)+logy+c=0$

Answer 1

The correct option is D

$sin{h}^{-1}\left(\frac{x}{y}\right)+logy+c=0$

$\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{si{n}^{2}x}{1+x^3}P=\frac{3x^2}{1+x^3}\Rightarrow I.F.=e^{\int pdx}=e^{log(1+x^3)=1+x^3}$
Thus the solution is
$y.(1+x^3)=\int \frac{si{n}^{2}x}{1+x^3}(1+x^3)dx=\int \frac{1-cos2x}{2}dx\Rightarrow y(1+x^3)=\frac{1}{2}x-\frac{sin2x}{4}+c.$