Multivariable Integration

An interesting one !

We have, $\displaystyle f(x)=\sin x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin x+m\cos x)f(m)dm$

Differentiating twice we have, $\displaystyle f''(x) = -f(x)$ which is popularly known as Van Der Pol's Equation .

The solution to the equation is given by , $\displaystyle f(x)=k_1\sin x+k_2\cos x$

So we have two forms of $f(x)$ namely,

$\displaystyle \begin{cases} f(x)=\sin x+\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\sin x+m\cos x)f(m)dm \space \space \text{ (*)} \\ f(x)=k_1\sin x+k_2\cos x \space \space \text{ (**)}\end{cases}$

From $(**)$ we have $f(0)=k_2$ & $f(\frac{\pi}{2}) = k_1$ ,

Now from $(*)$ we have , $\displaystyle f(0) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} xf(x)dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x(k_1\sin x+k_2\cos x)dx = 2k_1\int_{0}^{\frac{\pi}{2}} x\sin x dx =2k_1$

Since $f(0)=k_2$ , $\displaystyle k_2=2k_1$

Similarly we have, $\displaystyle k_1=1+2k_2$ from $f(\frac{\pi}{2})$

Solving we have , $\displaystyle \begin{cases} k_1=-\frac{1}{3} \\ k_2=-\frac{2}{3} \end{cases}$

Therefore , $\displaystyle \int_{0}^{\frac{\pi}{2}} f(x) dx = -\frac{1}{3}-\frac{2}{3} = -1$

$\text{Note :}$

Solution to Van-Der-Pol's Equation is as follows :

$\displaystyle \frac{d^2 y}{dx^2}+y=0\implies (D^2+1)y=0$ where $'D'$ is the differential operator

According to the rules if the roots of the polynomial in $D$ are imaginary and of the form $\alpha\pm i\beta$ then the solution is givcen by, $\displaystyle y= e^{-\alpha x} (c_1 cos(\beta x)+c_2\sin(\beta x))$

Here the solutions are $\pm i$ so the solution of the DE is,

$\displaystyle y= c_1\cos x+c_2\sin x$