Question

The solution of $\frac{dy}{dx}-\frac{\tan y}{1+x}=(1+x)e^x\sec y$ is

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

Answer 1

The correct option is A $\frac{\sin y}{1+x}=e^x+c$

$\frac{dy}{dx}-\frac{tany}{1+x}=(1+x)e^{x}secy$

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

Multiplying both sides by $\cos y$

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

Now, Let $t=\sin y$

$dt=\cos ydy$

Substituting the values, the equation now is

$\frac{dt}{dx}-\frac{t}{1+x}=(1+x)e^x$

Here we have a first order differential equation of the form

$\frac{dt}{dx}+Pt=Q$

where $P=\frac{-1}{1+x}andQ=(1+x)e^x$

Let's find the I.F

$I.F=e^{\int Pdx}$

$=e^{\int \frac{-1}{1+x}dx}$

$=e^{-log(1+x)}$

$=e^{log(1+x{)}^{-1}}$

$=(1+x{)}^{-1}$

$t\times I.F=\int (Q\times I.F)dx+C$

$t(1+x{)}^{-1}=\int (1+x\left)e^x\right(1+x{)}^{-1}dx+C$

$t(1+x{)}^{-1}=\int e^{x}dx+C$

$t(1+x{)}^{-1}=e^x+C$

$\frac{t}{1+x}=e^x+C$

Putting the value of t

$\frac{\sin y}{1+x}=e^x+C$