Question

The family of curves satisfying the differential equation $\frac{dy}{dx}+\frac{1}{x}\sin 2y=x^3{\cos }^{2}y,$ is
(where $C$ is an arbitrary constant)

• A. $x^6+6x^2=C\tan y$
• B. $6x^2\tan y=x^6+C$
• C. $\sin 2y=x^3{\cos }^{2}y+C$
• D. $y^6=6y^2\tan x+C$

Answer 1

The correct option is B $6x^2\tan y=x^6+C$

We have,
$\frac{dy}{dx}+\frac{1}{x}\sin 2y=x^3{\cos }^{2}y$
Dividing both the sides by ${\cos }^{2}y,$ we get
${\sec }^{2}y\frac{dy}{dx}+\frac{2}{x}\tan y=x^3$

Assuming $\tan y=z$, we get
$\frac{dz}{dx}+\frac{2}{x}z=x^3$
which is a linear differential equation in $z.$
So, the integrating factor is
$I.F.=e^{\int \frac{2}{x}dx}=e^{2\ln x}=x^2$

Therefore, the solution of the D.E. is
$z{x}^2=\int x^3\times x^{2}dx+c\Rightarrow z{x}^2=\frac{x^6}{6}+c\Rightarrow 6x^2\tan y=x^6+6c\therefore 6x^2\tan y=x^6+C[C=6c]$