<s>[INST] Find $y(1)$ if $y(x)=x^3+\int_0^x \sin(x-t)y(t)dt$ 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 $y(1)$. 
 [/INST] 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 $y(1) = \frac{21}{20}$.
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