Question

Solve the differential equation $\cos (x+y)dy=dx$. Hence, find the particular solution for $x=0$ and $y=0$.

Answer 1

The given differential equation is

$\frac{dy}{dx}=\frac{1}{\cos (x+y)}$ .... (i)

Let $x+y=v$

Differentiating w.r.t. x, we get,

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

$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$

Substituting $\frac{dy}{dx}$ in equation (i), we get

$\frac{dv}{dx}-1=\frac{1}{\cos v}$

$\frac{dv}{dx}=\frac{1}{\cos v}+1$

$\frac{dv}{dx}=\frac{1+\cos v}{\cos v}$

Therefore on integrating, we get

$\int \frac{\cos vdv}{1+\cos v}=\int dx+c$

$\int \frac{1+\cos v}{1+\cos v}dv-\int \frac{1}{1+\cos v}dv=\int dx+c$

$\int 1dv-\int \frac{1}{2{\cos }^2\frac{v}{2}}dv=x+c$

$v-\frac{1}{2}\int {\sec }^2\frac{v}{2}dv=x+c$

$v-\tan \frac{v}{2}=x+c$

$(x+y)-\tan \left(\frac{x+y}{2}\right)=x+c$ .... (ii)

which is the required general solution of the given equation.

For $x=0$ and $y=0$, equation (ii) becomes,

$(0+0)-\tan \left(\frac{0}{2}\right)=0+c$

$\Rightarrow c=0$

Putting the value of $c$ is equation (ii), we get

$x+y-\tan \left(\frac{x+y}{2}\right)=x$

$\Rightarrow y=\tan \left(\frac{x+y}{2}\right)$