Find $f(x)$ if $f(x)=A\int_0^{\frac{\pi}2}(\sin x\cos t\cdot f(t))dt+\sin x$, where $A$ is a constant

Asked Jun 13 '21 at 12:57

Active Jun 13 '21 at 13:15

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Find $f(x)$ if $f(x)=A\int_0^{\frac{\pi}2}(\sin x\cos t\cdot f(t))dt+\sin x$, where $A$ is a constant.

$$f(x)=(A\int_0^{\frac{\pi}2}(\cos t\cdot f(t))dt+1)\sin x\\=B\sin x,$$ where $B=A\int_0^{\frac{\pi}2}(\cos t\cdot f(t))dt+1$ .... $(1)$

Also, $f(t)=B\sin t$ .... $(2)$

Putting $(2)$ in $(1)$,$$B=A\int_0^{\frac{\pi}2}(\cos t\cdot B\sin t)dt+1\\\implies B=\frac{AB}2\int_0^{\frac{\pi}2}\sin(2t)dt+1\\\implies B=AB+1 $$

Don't know if I am doing right and/or how to finish.

EDIT: @Nicolas asked to do double derivative. So, here is my attempt:

$$f(x)=B\sin x\\f'(x)=B\cos x\\f''(x)=-B\sin x\\\implies f^{"}(x)=-f(x)$$

I must add that solving second order differential equations is not in the syllabus.

edited Jun 13 '21 at 13:15

asked Jun 13 '21 at 12:57

aarbee

Have you tried to differentiate $f$ twice? – Nicolas Jun 13 '21 at 13:00
@Nicolas Thanks. I have made an edit. – aarbee Jun 13 '21 at 13:16
Just to explain my idea: the standard theory of ODEs states that, necessarily, $f(x)=\alpha\sin(x)+\beta\cos(x)$ for some constants $\alpha,\beta$ that you may be able to find by injecting this ansatz for $f$ into the integral. – Nicolas Jun 13 '21 at 13:25
Statement is confusing. Find $f(x)$ and $f(t)=Bsin(t)$ seem at odds. – herb steinberg Jun 13 '21 at 15:22
@herbsteinberg Maybe my attempt is wrong. – aarbee Jun 13 '21 at 15:24
$\int_0^{\frac{\pi}{2}}sin(2t)dt=1$, not $2$, so$B=\frac{2}{2-A}$ What has this got to do with a differential equation? – herb steinberg Jun 13 '21 at 20:52
@herbsteinberg I wonder the same. I was just shooting in the dark. I don't know what the correct apporach is. – aarbee Jun 13 '21 at 20:56
Correct approach to what? It looks like the only question is find B. – herb steinberg Jun 13 '21 at 21:06

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