Question

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

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

Answer 1

Answer: (B)

$\frac{d y}{d x}=\frac{tan y}{x}+\frac{e^{\frac{1}{x^{2}}}}{x^{2}}sec ⁡ y$
$cos y\frac{d y}{d x}-\frac{sin ⁡ y}{x}=\frac{e^{\frac{1}{x^{2}}}}{x^{2}}$
Let, $sin y=t$
$cos y\frac{d y}{d x}=\frac{d t}{d x}$
$\Rightarrow \frac{d t}{d x}-\frac{t}{x}=\frac{e^{\frac{1}{x^{2}}}}{x^{2}}$
I.F. $=e^{- \displaystyle \int \frac{1}{x} d x}=e^{ln \left(\frac{1}{x}\right)}=\frac{1}{x}$
$\Rightarrow \frac{t}{x}=\displaystyle \int \frac{e^{\frac{1}{x^{2}}}}{x^{3}}dx$
$\Rightarrow \frac{t}{x}=\frac{- 1}{2}\displaystyle \int \left(\frac{- 2}{x^{3}}\right)e^{\frac{1}{x^{2}}}dx$
$\Rightarrow \frac{sin y}{x}=\left(\frac{- 1}{2}\right)e^{\frac{1}{x^{2}}}+C$