Question

The solution of the differential equation $xdy=\left(tany+\frac{e^{1/x^2}}{x}secy\right)dx$ is (where $C$ is the constant of integration)

• A. $siny=e^{\frac{1}{x^2}}+C$
• B. $\frac{2siny}{x}+e^{\frac{1}{x^2}}=C$
• C. $\frac{siny}{x}-e^{\frac{1}{x^2}}=C$
• D. $siny-x{e}^{\frac{1}{x^2}}=C$

Answer 1

Answer: (B)

$\frac{dy}{dx}=\frac{tany}{x}+\frac{e^{\frac{1}{x^2}}}{x^2}secy$
$cosy\frac{dy}{dx}-\frac{siny}{x}=\frac{e^{\frac{1}{x^2}}}{x^2}$
Let, $siny=t$
$cosy\frac{dy}{dx}=\frac{dt}{dx}$
$\Rightarrow \frac{dt}{dx}-\frac{t}{x}=\frac{e^{\frac{1}{x^2}}}{x^2}$
I.F. $=e^{-\int \frac{1}{x}dx}=e^{ln\left(\frac{1}{x}\right)}=\frac{1}{x}$
$\Rightarrow \frac{t}{x}=\int \frac{e^{\frac{1}{x^2}}}{x^3}dx$
$\Rightarrow \frac{t}{x}=\frac{-1}{2}\int \left(\frac{-2}{x^3}\right)e^{\frac{1}{x^2}}dx$
$\Rightarrow \frac{siny}{x}=\left(\frac{-1}{2}\right)e^{\frac{1}{x^2}}+C$