Value of $f(0)$ in differential equation.

Question

If $f(\pi) = 2$ and $\displaystyle \int^{\pi}_{0}\bigg[f(x)+f''(x)\bigg]\sin(x)dx=5.$

Then value of $f(0).$

Given that function $f(x)$ is continuous in $[0,\pi]$

Try: $$\int \bigg[f(x)+f''(x)\bigg]\sin xdx$$

$$ = \int f(x)\sin xdx+\sin xf'(x)+\int \cos xf'(x)dx$$

$$ = \int f(x)\sin xdx+\sin xf'(x)+\cos xf(x)-\int \sin x f(x)dx$$

So $$\int^{\pi}_{0}\bigg[f(x)+f''(x)\bigg]\sin(x)dx = \sin xf'(x)+\cos x f(x) = 5.$$

So $$\sin xf'(x)+\cos xf(x) = 5$$

Could some help me how i solve above differential equation. Thanks

Answer 1

Integrate by part twice. $\int_0^{\pi} f''(x)\sin\, x \, dx= -\int_0^{\pi} f'(x)\cos\, x \, dx=f(0) -f(\pi)(-1)-\int_0^{\pi} f(x)\sin\, x \, dx$ so we get $5=f(0) +f(\pi)$. Hence $f(0)=3$.

Answer 2

In fact, you have obtained that $$f'(x)\sin x+f(x)\cos x|_{0}^{\pi}=5$$which means that $$-f(\pi)+f(0)=5$$from which we obtain$$f(0)=7$$