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??

  • $\begingroup$ Your method works when $r$ is a constant. The equation has variable coefficients so the solution you have obtained is not correct. $\endgroup$ – Kavi Rama Murthy Oct 23 '18 at 8:15
  • $\begingroup$ Then, solving this equation should use another method?? Is there a "name" for that method so that I can try to figure it out.\ $\endgroup$ – Jason Ng Oct 23 '18 at 8:21
  • $\begingroup$ It would be slightly more accurate to integrate the instant frequency to get an approximate phase, with $t=\sinh(u)$ one gets $$\phi(t)=\int\sqrt{1+t^2}dt=\int\cosh^2(u)du=\tfrac12t\sqrt{1+t^2}+\tfrac12\text{Arsinh}(t)$$ and $y(t)\approx A\cos(\phi(t))+B\sin(\phi(t))$, again with $A=1$, $B=0$. $\endgroup$ – LutzL Oct 23 '18 at 13:48

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 $s\in[0,1]$.

  • $\begingroup$ Thank you very much. I need to spend more time to understand and try myself. $\endgroup$ – Jason Ng Oct 23 '18 at 15:39
  • $\begingroup$ Yet, I saw a other methods: integrate the function directly: $$y"(t)+y(t)+t^2y(t)=0$$ $$y'(t)-y'(0)+\int_{0}^{t}y(t) dt + \int_{0}^{t} t^2y(t) dt =0 \quad y'(0)=0$$ Then, integrate again and have $$y(t)-y(0)-\int_{0}^{t}\int_{0}{t}y(s) dsdt + \int_{0}^{t}\int_{0}^{t} s^2y(s) dsdt =0$$ $$y(t)-y(0)-\int_{0}^{t}(t-s)y(s) ds + \int_{0}^{t}\int_{0}^{t} s^2y(s) dsdt =0$$ And where I was stuck is the third integral $\int_{0}^{t}\int_{0}^{t} s^2y(s) dsdt$. I don't know how to simplify and continue. And by the way, it seems this method cannot reach the answer there is no any trigo function $\endgroup$ – Jason Ng Oct 23 '18 at 15:39
  • 1
    $\begingroup$ This is essentially the Taylor formula with integral remainder term, $$y(t)=y(0)+y'(0)t+\int_0^t(t-s)y''(s)\,ds.$$ Now insert $y''(s)=-(1+s^2)y(s)$. If one does the limits carefully you get this also from the second approach for $ω\to 0$ as then $\sin(ωx)/ω\to x$. $\endgroup$ – LutzL Oct 23 '18 at 15:56
  • 1
    $\begingroup$ You make a parametrization of the solution with two unknown functions, but the solution is only one function. Thus you have one functional relation of the coefficient functions for free that you can use to simplify the derivatives. Or you argue from the POV of the first order system so that then using the fundamental matrix/solution $$\pmatrix{y\\y'}=\pmatrix{y_1&y_2\\y_1'&y_2'}\pmatrix{c_1\\c_2}.$$ $\endgroup$ – LutzL Oct 25 '18 at 16:49
  • 1
    $\begingroup$ See also Method of Variation of Parameters - Assigning zero works? $\endgroup$ – LutzL Oct 25 '18 at 16:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.