Question

The solution of the differential equation $\frac{dy}{dx}=\frac{1}{xy[x^2\sin y^2+1]}$ is
(where $C$ is integration constant)

• A. $x^2(\cos y^2-\sin y^2-C{e}^{-y^2})=2$
• B. $y^2(\cos x^2-\sin y^2-C{e}^{-y^2})=2$
• C. $x^2(\cos y^2-\sin y^2-e^{-y^2})=4C$
• D. $x(\cos y^2-\sin y^2-e^{-y})=4C$

Answer 1

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

$\frac{dy}{dx}=\frac{1}{xy[x^2\sin y^2+1]}$
$\Rightarrow \frac{dx}{dy}=xy[x^2\sin y^2+1]$
$\Rightarrow \frac{1}{x^3}\frac{dx}{dy}-\frac{1}{x^2}y=y\sin y^2$
Putting $-1/x^2=u,$ the above equation can be written as
$\frac{du}{dy}+2uy=2y\sin y^2$
On comparing with $\frac{du}{dx}+Pu=Q,$
We get, $P=2y$ and $Q=2y\sin y^2$

Now, the I.F.$=e^{y^2}$
$\therefore$ Solution is $u{e}^{y^2}=\int 2y\sin y^2{e}^{y^2}dy+C$
$=\int \left(\sin t\right)e^{t}dt+C$
$=\frac{1}{2}{e}^{y^2}(\sin y^2-\cos y^2)+c$
$\Rightarrow 2u=(\sin y^2-\cos y^2)+C{e}^{-y^2},C=2c$
$\Rightarrow 2=x^2[\cos y^2-\sin y^2-C{e}^{-y^2}]$