Question

The solution of the differential equation
$\frac{dy}{dx}=\frac{1}{xy[x^2\sin y^2+1]}$ is:

• A. $x^2(\cos y^2-\sin y^2-2c{e}^{-y^2})=2$
• B. $y^2(\cos x^2-\sin y^2-2c{e}^{-y^2})=2$
• C. $x^2(\cos x^2-\sin y^2-c{e}^{-y^2})=4$
• D. none of these

Answer 1

The correct option is A $x^2(\cos y^2-\sin y^2-2c{e}^{-y^2})=2$

The given differential equation can be written as
$\frac{dx}{dy}=xy(x^2\sin y^2+1)\Rightarrow \frac{1}{x^3}\frac{dx}{dy}-\frac{1}{x^2}y=y\sin y^2$
This equation is reducible to linear equation.
So putting $v=-\frac{1}{x^2}$
$\therefore \frac{dv}{dy}+2vy=2y\sin y^2$
The integrating factor of this equation is $e^{y^2}$
So the required equation is
$v{e}^{y^2}=\int 2y\sin y^2{e}^{y^2}dy+c=-\frac{1}{2}{e}^{y^2}\left(\sin y^2-\cos y^2\right)+c\Rightarrow 2v=\left(\sin y^2-\cos y^2\right)+e^{-y^2}\Rightarrow 2=x^2(\cos y^2-\sin y^2-2c{e}^{-y^2})$