Find $y(1)$ if $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt$

Question

Consider the integral equation $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt, x\in[0,\pi].$ Then the value of y(1) is

$1. \;\;19/20 \\ 2. \;\;1\\ 3. \;\;17/20\\ 4. \;\;21/20 $

My Attempt: $y(x)=x^3+\int_0^x \sin(x-t) \ y(t)dt \\ \Rightarrow y(x)=x^3+\int_0^x(\sin x \cos t-\cos x \sin t)\ y(t)dt \\ \Rightarrow y(x)=x^3+\sin x\int_0^x \cos t\ y(t)dt -\cos x\int_0^x \sin t\ y(t)dt................(1)\\ y'(x)=3x^2+ \sin x \cos x\ y(x)+\cos x\int_0^x \cos t\ y(t)dt-\cos x\sin x \ y(x)+\sin x\int_0^x \sin t\ y(t)dt \\ y'(x)=3x^2+\cos x\int_0^x \cos t \ y(t)dt+\sin x\int_0^x \sin t\ y(t)dt\\ y''(x)=6x+\cos x(\cos x\ y(x))-\sin x\int_0^x \cos t\ y(t)dt+\sin x(\sin x\ y(x))+\cos x\int_0^x \sin t\ y(t)dt \\ y''(x)=6x+(\cos^2x+\sin^2x)\ y(x)-(\sin x\int_0^x \cos t \ y(t)dt-\cos x\int_0^x \sin t \ y(t)dt) \\ \\ y''(x)=6x+y(x)-(y(x)-3x^2)\\ y''(x)=6x+3x^2\\$ now integrating twice we get $$y(x)=x^3+ \frac{x^4}{4}+c_1x+c_2 $$

Since $y(0)=0$, we get $c_2=0$.

I don't know how to solve further to find $y(1)$.

Answer 1

In the next-to-last line, you should have $y(x) - x^3$, so that $y(x) = x^3 + \frac{x^5}{20} + c_1x + c_2$. Then since line 5 implies $y'(0) = 0$, both $c_1$ and $c_2$ are zero, so that $y(x) = x^3 + \frac{x^5}{20}$, and $y(1) = \frac{21}{20}$.

Answer 2

If your 5th line from the top were correct $$y'(x) = 3x^2 +\cos x \int_0 ^x \cos t \,y(t) \mathrm{d}t + \sin x \int_0 ^x \sin t \, y(t) \mathrm{d} t$$ would it not imply $y'(0) = 0$, and hence $c_1 = 0$? That would though yield $y(1) = \frac{5}{4}$

Answer 3

With Laplace Transform:

$$\mathcal{L}_x\left[y(x)=x^3+\int_0^x \sin (x-t) y(t) \, dt\right](s)$$

$$\mathcal{L}_x[y(x)](s)=\frac{6}{s^4}+\frac{\mathcal{L}_x[y(x)](s)}{1+s^2}$$

$$\mathcal{L}_x[y(x)](s)=\frac{6 \left(1+s^2\right)}{s^6}$$

$$\mathcal{L}_s^{-1}\left[\mathcal{L}_x[y(x)](s)=\frac{6 \left(1+s^2\right)}{s^6}\right](x)$$

$$y(x)=\frac{1}{20} \left(20 x^3+x^5\right)$$

$$y(1)=\frac{21}{20}$$