2.3.6.2 second order linear constant coeff

\[\begin {aligned} x^{\prime \prime }+x&=\cos \left (t \right )\\ x \left (0\right ) &= 0\\ x^{\prime }\left (0\right ) &= 1\\ \end {aligned}\]

Entering second order linear constant coefficient ode solverThis is second order non-homogeneous ODE. In standard form the ODE is

\[ A x''(t) + B x'(t) + C x(t) = f(t) \]

Where \(A=1, B=0, C=1, f(t)=\cos \left (t \right )\). Let the solution be

\[ x = x_h + x_p \]

Where \(x_h\) is the solution to the homogeneous ODE

\begin{align*} A x''(t) + B x'(t) + C x(t) &= 0 \end{align*}

And \(x_p\) is a particular solution to the non-homogeneous ODE

\begin{align*} A x''(t) + B x'(t) + C x(t) &= f(t) \end{align*}

Where \(x_h\) is the solution to

\begin{align*} x^{\prime \prime }+x = 0 \end{align*}

This is second order with constant coefficients homogeneous ODE. In standard form the ODE is

\[ A x''(t) + B x'(t) + C x(t) = 0 \]

Where in the above \(A=1, B=0, C=1\). Let the solution be \(x=e^{\lambda t}\). Substituting this into the ODE gives

\[ \lambda ^{2} {\mathrm e}^{\lambda t}+{\mathrm e}^{\lambda t} = 0 \tag {1} \]

Since exponential function is never zero, then dividing Eq(2) throughout by \(e^{\lambda t}\) gives

\[ \lambda ^{2}+1 = 0 \tag {2} \]

Equation (2) is the characteristic equation of the ODE. Its roots determine the general solution form. Using the quadratic formula the roots are

\[ \lambda _{1,2} = \frac {-B}{2 A} \pm \frac {1}{2 A} \sqrt {B^2 - 4 A C} \]

Substituting \(A=1, B=0, C=1\) into the above gives

\begin{align*} \lambda _{1,2} &= \frac {0}{(2) \left (1\right )} \pm \frac {1}{(2) \left (1\right )} \sqrt {0^2 - (4) \left (1\right )\left (1\right )}\\ &= \pm i \end{align*}

Hence

\begin{align*} \lambda _1 &= + i\\ \lambda _2 &= - i \end{align*}

Which simplifies to

\begin{gather*} \begin {aligned} \lambda _1 &= i\\ \lambda _2 &= -i \end {aligned} \end{gather*}

Since the roots are complex conjugate of each others, then let the roots be

\begin{align*} \lambda _{1,2} &= \alpha \pm i \beta \end{align*}

Where \(\alpha =0\) and \(\beta =1\). Therefore the final solution using Euler relation can be written as

\begin{align*} x &= e^{\alpha t} \left ( c_1 \cos (\beta t) + c_2 \sin (\beta t) \right ) \end{align*}

Which becomes

\begin{align*} x &= e^{0}\left (c_1 \cos \left (t \right )+c_2 \sin \left (t \right )\right ) \end{align*}

Or

\begin{align*} x &= c_1 \cos \left (t \right )+c_2 \sin \left (t \right ) \end{align*}

Therefore the homogeneous solution \(x_h\) is

\begin{align*} x_h &= c_1 \cos \left (t \right )+c_2 \sin \left (t \right ) \end{align*}

The particular solution is found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ \cos \left (t \right ) \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{\cos \left (t \right ), \sin \left (t \right )\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \{\cos \left (t \right ), \sin \left (t \right )\} \]

Since \(\cos \left (t \right )\) is duplicated in the UC_set, then this basis is multiplied by extra \(t\). The UC_set becomes

\[ [\{t \sin \left (t \right ), \cos \left (t \right ) t\}] \]

Since there was duplication between the basis functions in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis function in the above updated UC_set.

\begin{align*} x_p &= A_{1} t \sin \left (t \right )+A_{2} \cos \left (t \right ) t \end{align*}

The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(x_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\begin{align*} 2 A_{1} \cos \left (t \right )-2 A_{2} \sin \left (t \right ) = \cos \left (t \right ) \end{align*}

Solving for the unknowns by comparing coefficients results in

\[ \left [A_{1} = {\frac {1}{2}}, A_{2} = 0\right ] \]

Substituting the above back in the above trial solution \(x_p\), gives the particular solution

\begin{align*} x_p &= \frac {t \sin \left (t \right )}{2} \end{align*}

Therefore the general solution is

\begin{gather*} \begin {aligned} x &= x_h + x_p \\ &= \left (c_1 \cos \left (t \right )+c_2 \sin \left (t \right )\right ) + \left (\frac {t \sin \left (t \right )}{2}\right ) \end {aligned} \end{gather*}

Applying the initial conditions \([x \left (0\right ) = 0, x^{\prime }\left (0\right ) = 1]\), the solution becomes

\begin{align*} x &= \frac {\left (t +2\right ) \sin \left (t \right )}{2} \\ \end{align*}

\(x = \frac {\left (t +2\right ) \sin \left (t \right )}{2}\)	Direction field