The Initial Value problem $$y’’+y=0,y(0)=1,y’(0)=0$$ is equivalent to integral equation
$(A)$. $y(x)=1+\int_0^x(t-x)y(t)dt$
$( B)$. $y(x)=1+\int_0^x(t+x)y(t)dt$
$ (C)$. $y(x)=1+\int_0^x(tx)y(t)dt$
$(D)$. $y(x)=1+\int_0^x(x +t)y(t)dt$
If I assume $y’’=u(x) $ then by integration from $0$ to $x$ I got $$y’-y’(0)=\int_0^x u(t)dy$$ so that $$y’=\int_0^x u(t)dt$$ Again integration I got $$y=1+\int_0^x(x-t)u(t)dt$$ which i put in given differential equation I got $$u(x)=-1+\int_0^x(t-x)u(t)dt$$ But no option matches . Where I did mistake? Thank you.