Question

Solve the differential equation: $\frac{dy}{dx}-\frac{\tan y}{1+x}=(1+x)e^x\sec y$

• A. $\sin y=(e^x+c)(1+x)$
• B. $\sin y=(e^x+c)(1-x)$
• C. $\cos y=(e^x+c)(1+x)$
• D. $\cos y=(e^x+c)(1-x)$

Answer 1

The correct option is A $\sin y=(e^x+c)(1+x)$

$\frac{dy}{dx}-\frac{\tan y}{1+x}=(1+x)e^x\sec y\Rightarrow \cos y\frac{dy}{dx}-\frac{\sin y}{1+x}=(1+x)e^x$
Put $\sin y=v\Rightarrow \cos ydy=dx$
$\therefore \frac{dv}{dx}-\frac{v}{1+x}=(1+x)e^x$ ...(1)
Here $P=-\frac{1}{1+x}\Rightarrow \int Pdx=-\int \frac{1}{1+x}dx=-\log (1+x)=\log \left(\frac{1}{1+x}\right)$
$\therefore I.F.=e^{\log \left(\frac{1}{1+x}\right)}=\frac{1}{1+x}$
Multiplying (1) by $I.F.$ we get
$\frac{1}{1+x}\frac{dv}{dx}-\frac{v}{{(1+x)}^2}=e^x$
Integrating both sides we get
$\frac{v}{1+x}=\int e^{x}dx+c=e^x+c\Rightarrow \sin y=(e^x+c)(1+x)$