Question is to :
Form an integral equation corresponding to the initial value problem $$\frac{d^2y}{dx^2}+y=0 ~; ~ x>0$$ with initial conditions $y(0)=1$ and $y'(0)=0$
What i have tried so far is :
Put $$\frac{d^2y}{dx^2}=\psi(x)$$
I use fundamental theorem of calculus $$\int_a^b f(t)dt=F(b)-F(a)$$ where $F$ is anti derivative of $f$.
In this case we would have
$$\int_0^x\psi(t)dt=\int_0^x \frac{d^2y}{dt^2}dt=F(x)-F(0)=\frac{dy}{dx}-y'(0)$$
i.e., $$\frac{dy}{dx}=\int_0^x\psi(t)dt+y'(0)$$
i.e., $$\frac{dy}{dx}=\int_0^x\psi(t)dt$$
But i am confused how to get $y$ from this (though i am sure i should use same fundamental theorem of calculus)....
If I got $y(x)$ then I would just substitute that in differential equations to get integral equation.
Please do not write whole answer at once... I would like to get some hints and try to learn something before concluding for an answer.
Kindly co operate with me..
Thank you. :)
Why not use the following relation? $$\psi (x)=\frac{d^2y}{dx^2}=-y$$
Letting $D$ be differential operation, we have $$(D^2y)+y=0\Rightarrow (D^2+1)y=0\Rightarrow (D-i)(D+i)y=0$$ This means that letting $z=(D-i)y=Dy-iy=y^\prime-iy$, then $(D+i)z=0$.
Namely, $Dz+iz=0\Rightarrow z^\prime+iz=0.$
Do you think you can continue?