Question

The solution of ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$ is:

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

Answer 1

The correct option is C $y{e}^{\tan x}=(\tan x-1)e^{\tan x}+c$

${\cos }^{2}x\frac{dy}{dx}+y=\tan x$

$\Rightarrow \frac{dy}{dx}+\frac{y}{{\cos }^{2}x}=\frac{\tan x}{{\cos }^{2}x}$

$\Rightarrow \frac{dy}{dx}+y{\sec }^{2}x=\tan x{\sec }^{2}x$

Here $P={\sec }^{2}x$ and $Q=\tan x{\sec }^{2}x$

$I.F=e^{{\int }^{}Pdx}=e^{{\int }^{}{\sec }^{2}xdx}=e^{\tan x}$

The solution of given Differential equation is

$y\times I.F={\int }^{}Q\times I.Fdx+c$

$\Rightarrow y\times e^{\tan x}={\int }^{}\tan x{\sec }^{2}x\times e^{\tan x}dx+c$

putting $\tan x=t$

$\Rightarrow {\sec }^{2}xdx=dt$

$\Rightarrow y{e}^{\tan x}={\int }^{}t{e}^{t}dt+c$

$\Rightarrow y{e}^{\tan x}=t{e}^t-e^t+c$

$\Rightarrow y{e}^{\tan x}=e^t\left(t-1\right)+c$

$\Rightarrow y{e}^{\tan x}=e^{\tan x}\left(\tan x-1\right)+c$