Given an IVP $$y''(t)+(1+t^2)y=0 \quad y(0)=1, y'(0)=0$$ (a) Show that it is equivalent to $$y(t)=\cos(t)+\int_{0}^{t} s^2\sin(s-t)y(s) ds$$
My methods: For the reverse part, I am able to deduce that the given integral equation is the solution of the IVP by direct differentiation.
However, for the front part, I am not able to finish it. First, characteristic equation $r^2=-(1+t^2)$ gives me two complex roots. Hence, $$y(t)=A\cos(\sqrt{1+t^2}t) +B\sin(\sqrt{1+t^2}t)$$ Use the IVP $y(0)=1$, $A=1$ and $B=0$. Hence, $$y(t)=\cos(\sqrt{1+t^2}t)$$ Then, how can I continue??
Apply the formula for the variation of constants to the equation $$ y''(t)+y(t)=f(t)~~\text{ where }~~ f(t)=-t^2y(t). $$ There $$ y(t)=c_1(t)\cos(t)+c_2(t)\sin(t)\\ y'(t)=-c_1(t)\sin(t)+c_2(t)\cos(t) $$ with the conditions $$ \left.\begin{aligned} c_1'(t)\cos(t)+c_2'(t)\sin(t)&=0\\ -c_1'(t)\sin(t)+c_2'(t)\cos(t)&=f(t) \end{aligned}\right\} \implies \left\{\begin{aligned} c_1'(t)&=-\sin(t)f(t)\\ c_2'(t)&=\cos(t)f(t) \end{aligned}\right. $$ so that $$ \left.\begin{aligned} c_1(t)&=c_1(0)-\int_0^t\sin(s)f(s)\,ds\\ c_2(t)&=c_2(0)+\int_0^t\cos(s)f(s)\,ds \end{aligned}\right\} $$ Combining these formulas, inserting the initial condition at $t=0$ and applying trigonometric identities leads then to the given formula. \begin{align} y(t)&=\left(1-\int_0^t\sin(s)f(s)\,ds\right)\cos(t)+\left(\int_0^t\cos(s)f(s)\,ds\right)\sin(t)\\ &=\cos(t)+\int_0^t[-\sin(s)\cos(t)+\cos(s)\sin(t)]f(s)\,ds\\ &=\cos(t)-\int_0^t\sin(t-s)\,s^2\,y(s)\,ds. \end{align}
In a similar way, if you take any other frequency $\omega$ and consider the function $$ g_t(s)=ω\cos(ω(t-s))y(s)+\sin(ω(t-s))y'(s) $$ you get the derivative $$ g_t'(s)=\sin(ω(t-s))(ω^2y(s)+y''(s))=\sin(ω(t-s))(ω^2-1-s^2)y(s) $$ Now integrating both sides from $0$ to $t$ gives $$ ωy(t)-\bigl[ω\cos(ωt)y(0)+\sin(ωt)y'(0)\bigr]=g_t(t)-g_t(0)=\int_0^t\sin(ω(t-s))(ω^2-1-s^2)y(s)\,ds. $$ With for example $ω=\frac54$ the factor $(ω^2-1-s^2)=(\frac9{16}-s^2)$ is balanced in value for $s\in[0,1]$.