Question

The equation of the curve passing through the origin and satisfying the equation $x\frac{dy}{dx}+\sin 2y=x^4{\cos }^{2}y$ is

• A. $x^4=6\tan y$
• B. $x^5=5\tan y$
• C. $x^6=6\tan y$
• D. $x^4=4\tan y$

Answer 1

The correct option is A $x^4=6\tan y$

$x\frac{dy}{dx}+\sin 2y=x^4{\cos }^{2}y$
Dividing by $x{\cos }^{2}y$, we get
${\sec }^{2}y\frac{dy}{dx}+\frac{2}{x}\tan y=x^3$
Put $\tan y=z\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{dz}{dx}$
$\therefore \frac{dz}{dx}+\left(\frac{2}{x}\right)z=x^3$, which is a linear differential equation.
$\text{I.F.}=\exp \left(\int \frac{2}{x}dx\right)$
$=e^{2\ln |x|}=e^{\ln x^2}=x^2$

The solution is $z\cdot x^2=\int (x^3\cdot x^2)dx$
$\Rightarrow x^2\tan y=\frac{x^6}{6}+C$
Given that the curve passses through the origin.
$\therefore 0=0+C\Rightarrow C=0$
Hence, equation of the curve is
$x^2\tan y=\frac{x^6}{6}$
$\Rightarrow x^4=6\tan y$