A second order DE is given:
$$u'' +\lambda^2u=g(t,u) \space , u(0)=0,u'(0)=1$$ where $\lambda>0$ for all $t\geq 0$.
The solution has to be verified which is:$$u(t)=\frac{\sin(\lambda t)}{\lambda}+\int_{0}^{t} \frac{\sin\lambda(t-s)}{\lambda}g(s,u(s))ds$$
My approach:
I figured out the $$y_c=c_1\cos(\lambda t)+c_2\sin(\lambda t)$$
and $$y_p=-cos(\lambda t) \int_{0}^{t} \frac{\sin(\lambda s)}{\lambda}g(s,u(s))ds+\sin(\lambda t)\int_{0}^{t} \frac{\cos(\lambda s)}{\lambda}g(s,u(s))ds$$
I used variation of parameter for the $y_p$ .
the general solution should be:
$$y=y_c+y_p$$
My questions are
1.If I plug in the initial conditions in only $y_c$ then I get $c_1=0$ and $c_2 =1/\lambda $ . Or should I have to plug in the initial conditions in the general one?
2 How to get rid of the $\space\space\sin(\lambda t)\int_{0}^{t} \frac{\cos\lambda(t-s)}{\lambda}g(s,u(s))ds$ part of the particular solution to get the desired solution ?
Best Answer
Consider $$g_t(s)=λ\cos(λ(s-t))y(s)-\sin(λ(s-t))y'(s).$$ Its derivative is $$ g_t'(s)=-\sin(λ(s-t))g(s,u(s)) $$ and thus by the fundamental theorem of calculus $$ λy(t)-[λ\cos(λt)y(0)+\sin(λt)y'(0)]=g_t(t)-g_t(0)=\int_0^tg_t'(s)\,ds =\int_0^t\sin(λ(t-s))g(s,u(s))\,ds. $$
Note that this is not a solution formula, but an integral equation that has the same solution as the initial ODE.