- #1
psie
- 266
- 32
- 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?
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?