### Differential equations: Linear second-order differential equations

### The Ansatz

In *General solution of a homogeneous linear second-order ODE with constant coefficients* we discussed how to solve a homogeneous linear second-order differential equation with constant coefficients. In light of *Linear structure of linear GDVs*, the solution of an arbitrary (inhomogeneous) linear second-order differential equation with constant coefficients, only requires that we find a single particular solution in addition to the homogeneous. After all, the general solution of an inhomogeneous differential equation can be written as \[ y = y_{\rm part} + y_{\rm hom} \] where #y_{\rm part}# is a particular solution of the equation and #y_{\rm hom}# is the general solution of the corresponding homogeneous equation.

*Later, at Variation of constants*, we will describe a method, based on the general homogeneous solution, for finding a particular solution. This method is laborious. This is one of the reasons to discuss the approach below. The current approach does not always provide a guarantee of success, but it is straightforward and is also applicable to other differential equations than the linear second order ODE's that are being studied here.

Ansatz method

Consider the linear differential equation of second order \[ a(t) \cdot y'' + b(t) \cdot y' + c(t) \cdot y = g(t) \] where #a#, #b#, #c#, and #g# are continuous functions.

In order to find a particular solution of the ODE we make up a conjecture regarding the form of a solution solution in terms of a function rule which includes parameters. Such a possible form of a solution is called an **Ansatz.** Filling in the Ansatz in the ODE provides a system of equations in the parameters. A solution of this system then gives values of the parameters which turn the Ansatz into a particular solution of the ODE.

The word comes from the German Ansatz but is used in many languages. It means "starting" or "effort". The Ansatz here thus consists of an assumption regarding the form of the solution. The method is also called *method of undetermined coefficients.*

It is of course possible that the system of equations, whose unknowns are the parameters of the Ansatz, has no solution. This means that the ODE has no solutions in the form suggested by the Ansatz.

We discuss four specific cases. It is not necessary to memorize the precise conditions. The aim is to given an impression of the solution strategy.

Ansatz examples for linear second-order ODEs

Consider a linear differential equation of second order \[ y'' + p \cdot y' + q\cdot y = g(t) \] where #p# and #q# are constants and #g# is a continuous function.

- If #g(t)# is a polynomial of degree #n\ge0# and #q\ne0#, then there is a particular solution #y_{\rm part}# which is a polynomial of degree #n#.
- If \(g(t)= g_0 \cdot \e^{k\cdot t}\), where #g_0# and #k# are constants, so that #k# is not a solution of the
*characteristic equation*#\lambda^2+p\cdot \lambda +q=0#, then there is a particular solution of the shape \[y_{\rm part}(t) = A\cdot \e^{k\cdot t}\] - If \(g(t)= g_0 \cdot \e^{k\cdot t}\), where #g_0# is a constant and #k# is a solution but not the only solution of the
*characteristic equation*#\lambda^2+p\cdot \lambda +q=0#, then there is a particular solution of the form \[y_{\rm part}(t) = A\cdot t\cdot \e^{k\cdot t}\] - If \(g(t)=g_0\cdot \cos(\beta\cdot t)+g_1 \cdot\sin(\beta\cdot t)\), where #g_0#, #g_1#, and #\beta# are constants with #\beta\ne0#, and if #p\ne0#, then there is a particular solution of the form \[ y_{\rm part}(x) = A \cdot\cos (\beta\cdot t) + B\cdot \sin (\beta\cdot t) \]

#y(t) =# #-2 t^2-10 t-24#

Because #g(t)=2\cdot t^2+2\cdot t# is a polynomial of degree #2#, we seek a particular solution of the form: \[y(t) = a_{2}\cdot t^2+a_{1}\cdot t+a_{0}\]

The first two derivatives of #y# are #y'(t) = 2\cdot a_{2}\cdot t+a_{1}# and #y''(t) = 2\cdot a_{2}#. If we substitute these function rules in the differential equation, we find:

\[\begin{array}{rcl}

y'' +2 y' -y&=&2\cdot t^2+2\cdot t \\

\color{blue}{\text{the differential equation}}&& \\

\left(2\cdot a_{2} \right) +2\cdot (2\cdot a_{2}\cdot t+a_{1}) -(a_{2}\cdot t^2+a_{1}\cdot t+a_{0}) &=& 2\cdot t^2+2\cdot t \\

\color{blue}{\text{function rules substituted}}&& \\

-a_{2}\cdot t^2+\left(4\cdot a_{2}-a_{1}\right)\cdot t+2\cdot a_{2}+2\cdot a_{1}-a_{0} &=& 2\cdot t^2+2\cdot t \\

\color{blue}{\text{rewritten in standard polynomial form}}&&

\end{array}\]

Polynomials are equal if and only if their coefficients are equal, so:

\[ -a_{2}=2 \qquad\quad 4\cdot a_{2}-a_{1}=2\qquad\quad 2\cdot a_{2}+2\cdot a_{1}-a_{0}=0\]

The solution of this system is:

\[ a_2 = -2 \qquad\quad a_1= -10 \qquad\quad a_0 = -24 \]

Therefore, a particular solution is

\[y(t) = -2 t^2-10 t-24\]

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