Question

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

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

Answer 1

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

Given differential equation is $\frac{dy}{dx}+\frac{\tan y}{1+x}=(1+x)e^x\sec y$

$\Rightarrow \frac{1}{\sec y}\frac{dy}{dx}+\frac{\tan y}{(1+x)\sec y}=(1+x)e^x$

$\Rightarrow \cos y\frac{dy}{dx}+\frac{\sin y}{1+x}=(1+x)e^x$

Substitute $\sin y=v\Rightarrow \cos ydy=dv$

$\therefore \frac{dv}{dx}+\frac{v}{1+x}=(1+x)e^x$ ...(1)

Here, $P=\frac{1}{1+x}\Rightarrow \int Pdx=\int \frac{1}{1+x}dx=\log (1+x)$

$\therefore I.F.=e^{\log (1+x)}=1+x$

Multiplying $1$ by $I.F.$, we get $(1+x)\frac{dv}{dx}+v={(1+x)}^2.e^x$

Integrating both sides

$(1+x)v=\int {(1+x)}^2.e^{x}dx+C$

$\Rightarrow \sin y=(1+x)(e^x+C)$