The Ansatz

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.

In the definition of characteristic equation for a linear second-order ODE with constant coefficients, we filled #y=\e^{\lambda\cdot t}# in the ODE in order to find a solution of that form for a value of #\lambda# that was yet to be determined. This is an example of an Ansatz.

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 \[ y_{\rm part}(x) = A \cdot\cos (\beta\cdot t) + B\cdot \sin (\beta\cdot t) \]

The solution method consists of substituting the Ansatz in the differential equation and solving each of the resulting equations for the parameters (like the coefficients of the polynomial in the first case and the parameters #A# and #B# to be found in the other two cases).

Compare the third case with the classification by the discriminant #D# of the characteristic polynomial in the general solution of a homogeneous linear second-order GDV with constant coefficients. Here, we must have #D\lt0#.

In the fourth case, the condition #p\ne0# prevents the Ansatz from coinciding with the homogeneous solution in the case #D\lt0#.

Of course, many other Ansatz's are possible. A frequently used example is an infinite series, around #t=0#, like #y(t) = \sum_{i=0}^\infty a_it^i#. The idea is then to derive a recurrence relation between the unknown constants #a_i#. The result will then have to be a convergent series on an open interval around #t=0#.

For example, for Laguerre's differential equation

\[t\cdot y''+(1-t)\cdot y'+\lambda\cdot y = 0\]

this Ansatz gives the solution

\[y=\sum_{i=0}^\infty a_it^i =a_0\cdot\left(1+ \sum_{i=0}^\infty \dfrac{-\lambda\cdot(-\lambda+1)\cdots(-\lambda+i-1)}{(i-1)!}\cdot\dfrac{t^i}{i!}\right)\]

If #\lambda # is a natural number, then this solution is a polynomial of degree #\lambda#.

We treat each of the four cases separately.

Case #g(t)=\sum_{i=0}^ng_it^i#: Write #y(t)=\sum_{i=0}^na_it^i#. Differentiation renders

\[\begin{array}{rclcl} y'&=& \sum_{j=0}^n j\cdot a_j t^{j-1}&=&\sum_{i=0}^{n-1}(i+1)\cdot a_{i+1}t^i\\ y''&=& \sum_{j=0}^nj\cdot(j-1)\cdot a_jt^{j-2}&=&\sum_{i=0}^{n-2}(i+1)(i+2)\cdot a_{i+2}t^i\end{array}\]

After substitution of all function rules, the ODE becomes

\[\left.\begin{aligned}& \sum_{i=0}^{n-2} \left( (i+1)(i+2)\cdot a_{i+2}+p\cdot(i+1)\cdot a_{i+1}+q\cdot a_i\right) \cdot t^i & \\ &+(p\cdot n\cdot a_n +q\cdot a_{n-1})\cdot t^{n-1}&\\ &+q\cdot a_nt^n &\end{aligned}\right\}= \sum_{i=0}^ng_it^i \]

Comparison of the leading coefficient (that is, the coefficient of #t^n# ) on both sides gives

\[a_n= \frac{g_n}{q}\]

If we compare the coefficients of #t^j# for #j=n-1, n-2, \ldots# sequentially, we always find a linear equation in #a_i# with coefficient #q# in which all #a_i# with #i\gt j# are known. These equations are soluble, and thus lead to a particular solution #y(t)=\sum_{i=0}^n a_it^i#.

Case #g(t)=g_0\cdot\e^{k\cdot t}# where #k# is not a solution of the characteristic equation: substituting #y(t) =A\cdot \e^{k\cdot t}# in the ODE yields, after dividing out the factor #\e^{k\cdot t}#,

\[A\left(k^2+p\cdot k + q\right)=g_0\]

The assumption that #k# is not a solution of the characteristic equation means #k^2+p\cdot k + q\ne0#, so we can solve the equation found: \[A =\dfrac{g_0}{k^2+p\cdot k + q}\]

Case #g(t)=g_0\cdot\e^{k\cdot t}# which #k# a solution but not the only solution of the characteristic equation: filling #y(t) =A\cdot t\cdot \e^{k\cdot t}# into the ODE gives, after dividing out the factor #\e^{k\cdot t}#,

\[A\left(k^2+p\cdot k + q\right)\cdot t + 2k\cdot A+p\cdot A =g_0\]

The assumption that #k# is a solution but not the only solution of the characteristic equation #\lambda^2+p\cdot \lambda +q=0# means that #k^2+p\cdot k + q=0# and that the discriminant #p^2-4q# is distinct from #0#, which implies that #p+2k\ne0#. In particular, we can solve the equation found: \[A =\dfrac{g_0}{p+2k}\]

Case #g(t)=g_0\cdot\cos(\beta\cdot t)+g_1\cdot\sin(\beta\cdot t)#: calculation of the derivatives of #y(t)= A \cdot\cos (\beta\cdot t) + B\cdot \sin (\beta\cdot t)# and filling the function rules in the ODE provides an equation for the coefficients of #\cos(\beta\cdot t)# and #\sin(\beta\cdot t)#:

\[\begin{array}{lcrcl}\text{for } \cos(\beta\cdot t):&\phantom{x}&(q-\beta^2)\cdot A +p\cdot\beta\cdot B&=&g_0\\ \text{for } \sin(\beta\cdot t):&\phantom{x}&-p\cdot\beta\cdot A +(q-\beta^2)\cdot B&=&g_1\end{array}\]

To show that this system of two equations with unknowns #A# and #B# has a unique solution, we use linear algebra. In matrix form the system can be written as

\[\begin{pmatrix}q-\beta^2& p\cdot \beta\\ -p\cdot \beta&q-\beta^2\end{pmatrix} \begin{pmatrix}A\\ B\end{pmatrix}=\begin{pmatrix}g_0\\ g_1\end{pmatrix}\]

Such a system with unknowns #A# and #B# has a unique solution as the determinant #(q-\beta^2)^2+4p^2\beta^2=q^2-2\beta\cdot q +\left(\beta^4+4p^2\cdot \beta^2\right)# of the #2\times2#-matrix is distinct from zero. However, if this determinant would be equal to zero, then #q# be a solution of a quadratic equation with discriminant #\left(-2\beta\right)^2-4\left(\beta^4+4p^2\cdot \beta^2\right)=-16p^2\cdot\beta^2#. Because this discriminant is negative (we use here #\beta\ne0# and #p\ne0)#, this does not occur.

Find a particular solution of the differential equation \[y'' +2 y' -y =2\cdot t^2+2\cdot t \]

#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

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

New example