Question

Solution of the differential equation $\frac{dy}{dx}=\sin (x+y)+\cos (x+y)$ is:

• A. $\log |1+\tan \frac{x+y}{2}|=y+c$
• B. $\log |2+\sec \frac{x+y}{2}|=x+c$
• C. $\log |1+\tan (x+y\left)\right|=y+c$
• D. none of these

Answer 1

Answer: (D)

none of these

Given,Differential equation as $\frac{dy}{dx}=sin(x+y)+cos(x+y)---->A$

Let $(x+y)=t$

On differentiating once

$\Rightarrow \frac{d}{dx}(x+y)=\frac{dt}{dx}$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dt}{dx}$

On subsequent substitutions equation $A$ becomes as follows.

$\frac{dt}{dx}-1=sint+cost$

$\Rightarrow \frac{dt}{dx}=(sint+(1+cost\left)\right)$

$\Rightarrow \frac{dt}{dx}=(2sin\frac{t}{2}cos\frac{t}{2}+(2co{s}^2\frac{t}{2}\left)\right)$

$\Rightarrow \frac{dt}{dx}=2co{s}^2\frac{t}{2}(tan\frac{t}{2}+1)$

$\Rightarrow \frac{2se{c}^2\frac{t}{2}}{(1+tan\frac{t}{2})}dt=dx$

on Integrating both sides

$\Rightarrow \int \frac{2se{c}^2\left(\frac{t}{2}\right)}{(1+tan\frac{t}{2})}=\int dx------>B$

let $(1+tan\frac{t}{2})=k$

on differentiating once

$\frac{d}{dt}(1+tan\frac{t}{2})=\frac{dk}{dt}$

$(0+se{c}^2(\frac{t}{2}\left)\right)dt=dk$

$\left(se{c}^2\right(\frac{t}{2}\left)\right)dt=dk---->C$

On substituting C in B

$A\to \int \frac{dk}{k}=\int dx$

$\Rightarrow lnk=x+c$

But $k=(1+tan\frac{t}{2})$

$\Rightarrow ln(1+tan\frac{t}{2})=x+c$

But $t=x+y$

$\Rightarrow ln(1+tan\frac{(x+y)}{2})=x+c$ is the required solution of given diferential equation.