Avoid unpleasant integrals in solving IVP

[psie]

Homework Statement

Solve $t^2y''-2ty'+2y=t^2\sin{t^4}, t>0$ with initial values $y(1)=2, y'(1)=5$.

Relevant Equations

See first two paragraphs below.

The formula I'm given is that the general solution to a linear inhomogeneous system $x'(t)=A(t)x(t)+b(t)$ is $x(t)=F(t)\int F^{-1}(t)b(t)dt$, where $F(t)$ is the fundamental matrix to the linear homogenous system (here $A(t)$ is an $n\times n$ matrix function and $b(t)$ and $n\times 1$ matrix function, both continuous in some interval $I\subset \mathbb R$).

Since a linear $n$th order ODE $y^{(n)}(t)+a_{n-1}(t)y^{(n-1)}(t)+\ldots +a_0(t)y(t)=f(t)$ can be reduced to a system, the corresponding solution is $y(t)=R_1 (t)\int K_n(t)f(t)dt$, where $R_1(t)$ is the first row of the fundamental matrix $F(t)$ and $K_n(t)$ the last column of the inverse of the fundamental matrix $F^{-1}(t)$.

So here we are given the linear, second order ODE $$t^2y''-2ty'+2y=t^2\sin{t^4}.\tag1$$The homogeneous equation is a so-called Euler equation, i.e. of the form $t^ny^{(n)}(t)+a_{n-1}t^{n-1}y^{(n-1)}(t)+\ldots+a_1ty'(t)+a_0y(t)=0$, where $a_{n-1},\ldots,a_0$ are constants (see Wikipedia). I will omit the details, but the general solution to the homogeneous equation of $(1)$ is $$y_h(t)=Ct+Dt^2.$$ From this we can construct the fundamental matrix and compute its inverse. It is $$F(t)=\begin{bmatrix}
t&t^2\\
1&2t
\end{bmatrix}\qquad F^{-1}(t)=\begin{bmatrix}
2/t&-1\\
-1/t^2&1/t
\end{bmatrix}.$$
So using the formula of the general solution to a linear $n$th order ODE, i.e. $y(t)=R_1 (t)\int K_n(t)f(t)dt$, we have $$y(t)=t\int (-\sin{t^4})dt+t^2\int \frac{\sin{t^4}}{t}dt.$$ We can define $G(t)+C=\int (-\sin{t^4})dt$ and $H(t)+D=\int \frac{\sin{t^4}}{t}dt$, and we get $$y(t)=Ct+Dt^2+G(t)t+H(t)t^2.$$ But here I'm stuck, i.e. I do not know how check the initial values and find the solution to the IVP. Is there a way to avoid having to compute the integrals?

 

[pasmith]

Do not use indefinite integrals here. Always fix a lower limit; the point at which the initial value is specified is convenient. This paritcular solution therefore vanishes at this point, leaving the coefficients of the homogenous solutions to satisfy the initial condition. I would therefore define $$\begin{split} G(t) &= \int_1^t \sin u^4 \,du \\ H(t) &= \int_1^t \frac{\sin u^4}{u} \,du \end{split}$$ so that $G(1) = H(1) = 0$. Then $$\begin{split} y(x) &= Ct + Dt^2 - tG(t) + t^2H(t) \\ y'(x) &= C + 2Dt - G(t) - t\sin t^4 + 2tH(t) + t\sin t^4 \\ &= C + 2Dt - G(t) + 2tH(t).\end{split}$$

 

[psie]

Is there a formula (I would assume in terms of the fundamental matrix) for the particular solution of an $n$th order linear ODE with any initial values?

Following my reasoning above, I would assume it would read $$y_p(t)=R_1 (t)\int_{t_0}^t K_n(u)f(u)du,$$ where $R_1$ is the first row of the fundamental matrix and $K_n$ the last column of the inverse of the fundamental matrix, but I'm not sure.

 

[psie]

I'd be very grateful if someone could confirm the following.

In #1 I gave the solution to an inhomogeneous system, namely $$x(t)=F(t)\int (F(t))^{-1}b(t)dt.$$ Using definite integrals, i.e. we have some initial condition $x(t_0)=x_0$, the above solution reads $$x(t)=F(t)(F(t_0))^{-1}x_0+F(t)\int_{t_0}^t (F(u))^{-1}b(u)du.$$ Now, the solution to the scalar $n$th order linear ODE should just be the first component of $x(t)$, meaning only the first component of $F(t)(F(t_0))^{-1}x_0$ and first component of $F(t)\int_{t_0}^t (F(u))^{-1}b(u)du$. So the solution $y(t)$ of the scalar $n$th order linear ODE, with initial values $(y(t_0),\ldots,y^{(n-1)}(t_0))=x_0$, reads $$y(t)=P_1(t,t_0)x_0+R_1 (t)\int_{t_0}^t K_n(u)b(u)du,$$ where $P_1(t,t_0)$ is the first row of $F(t)(F(t_0))^{-1}$, $R_1(t)$ the first row of $F(t)$ and $K_n(u)$ the $n$th column of $(F(u))^{-1}$.