Variation of constants

Differential equations: Solution methods for linear-second order ODEs

Variation of constants

We show how a particular solution can be found from a linear differential equation of second order if we have a pair of linearly independent solutions $y_1$ and $y_2$ of the corresponding homogeneous equation.

Consider the differential equation $y''+p(t)\cdot y'+q(t)\cdot y=g(t)$ in the unknown function $y$ of $t$ , where $p$ , $q$ , and $g$ are continuous functions.

Suppose that $y_1$ and $y_2$ are solutions of the corresponding homogeneous equation with Wronskian $W$ distinct from $0$ . Let $c_1$ and $c_2$ be differentiable functions which are each determined, up a constant, by

$\eqs{c_1'(t) &=& -\frac{1}{W(t)}\cdot y_2(t)\cdot g(t)\cr c_2'(t) &=& \frac{1}{W(t)}\cdot y_1(t)\cdot g(t)}$

Then $u(t) = c_1(t)\cdot y_1(t)+c_2(t)\cdot y_2(t)$

is a particular solution of the original ODE.

Let $c_1$ and $c_2$ be as indicated. It is easy to verify that their derivatives satisfy the following equation:

$c_1'(t)\cdot y_1(t)+c_2'(t)\cdot y_2(t)=0$

The fact that $u(t)$ is a particular solution, means

$u''(t)+p(t)\cdot u'(t)+q(t)\cdot u(t)=g(t)$

To work out this equation in terms of $c_1(t)$ and $c_2(t)$ we use $u(t) = c_1(t)\cdot y_1(t)+c_2(t)\cdot y_2(t)$ and the above equation with $c_1'$ , $c_2'$ . First we calculate $u'(t)$ and $u''(t)$ :

$\begin{array}{rcl}u'(t) &=& \dfrac{\dd}{\dd t}\left(c_1(t)\cdot y_1(t)+c_2(t)\cdot y_2(t)\right)\\ &=&c_1'(t)\cdot y_1(t)+c_2'(t)\cdot y_2(t)+c_1(t)\cdot y_1'(t)+c_2(t)\cdot y_2'(t)\\ &&\phantom{xxx}\color{blue}{\text{sum and product rule for differentiation}}\\ &=&c_1(t)\cdot y_1'(t)+c_2(t)\cdot y_2'(t)\\ &&\phantom{xxx}\color{blue}{\text{above equation with }c_1, c_2'}\end{array}$

and

$\begin{array}{rcl}u''(t) &=& \dfrac{\dd}{\dd t}\left(c_1(t)\cdot y_1'(t)+c_2(t)\cdot y_2'(t)\right)\\ &=&c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)+c_1(t)\cdot y_1''(t)+c_2(t)\cdot y_2''(t)\\ &&\phantom{xxx}\color{blue}{\text{sum and product rule for differentiation}}\\ &=&c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)-c_1(t)\cdot p(t)\cdot y_1'(t)-c_1(t)\cdot q(t)\cdot y_1(t)\\&&-c_2(t)\cdot p(t)\cdot y_2'(t)-c_2(t)\cdot q(t)\cdot y_2(t)\\&&\phantom{xxx}\color{blue}{y_1\text{ and } y_2\text{ are solutions of the ODE}}\\ &=&c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)- p(t)\cdot u'(t)-q(t)\cdot u(t)\\ &&\phantom{xxx}\color{blue}{\text{definition of }u}\end{array}$

The left hand side of the ODE can be rewritten as

$u''(t)+ p(t)\cdot u'(t)+q(t)\cdot u(t)=c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)$

Thus, the statement that $u(t)$ is a solution of the inhomogeneous equation is equivalent to

$c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)=g(t)$

Together with the additional condition we have two linear equations in the unknowns $c_1'(t)$ and $c_2'(t)$ :

$\eqs{c_1'(t)\cdot y_1(t)+c_2'(t)\cdot y_2(t)&=&0\cr c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)&=&g(t)\cr}$

The values for $c_1'(t)$ and $c_2'(t)$ given in the theorem are the unique solution of this system of linear equations (this follows from Linear Algebra because the Wronskian of $y_1$ and $y_2$ is distinct from $0$ ).

Those familiar with linear algebra will be able to rewrite the pair of linear equations $\eqs{c_1'(t)\cdot y_1(t)+c_2'(t)\cdot y_2(t)&=&0\cr c_1'(t)\cdot y_1'(t)+c_2'(t)\cdot y_2'(t)&=&g(t)\cr}$ in matrix form, and find the solution by making use of the fact that the matrix $\begin{pmatrix}y_1(t)&y_2(t)\\ y_1'(t)&y_2'(t) \end {pmatrix}$ is invertible with inverse $\frac {1} {W (t)} \begin{pmatrix}y_2'(t)&-y_2(t)\\ -y_1'(t)&y_1(t) \end {pmatrix}$ .

The above method is called variation of constants.

The name variation of constants indicates that we replace the constants $c_1$ and $c_2$ of the solution $c_1\cdot y_1+c_2\cdot y_2$ of the homogeneous equation by functions of $t$ in order to find a particular solution.

The above method of finding a particular solution if the homogeneous solutions are known, always works, but it is cumbersome. Earlier we discussed the Ansatz method for quickly reaching a particular solution. Here too, it can be recommended in some common cases, to guess the nature of the functions $c_1$ and $c_2$ . We give a few examples.

Which solution is obtained by applying variation of constants to the linear second-order ODE
$y''+p\cdot y'+q\cdot y= g$
with constant coefficients $p$ , $q$ , and $g$ such that $q\ne0$ ?

$y=\frac{g}{q}$

We follow the variation of constants method to find the required solution. This solution has the form $c_1(t)\cdot y_1(t)+c_2(t)\cdot y_2(t)$

where $c_1$ and $c_2$ are each determined up to a constant by

$\eqs{c_1'(t) &=& -\frac{1}{W(t)}\cdot y_2(t)\cdot g\cr c_2'(t) &=& \frac{1}{W(t)}\cdot y_1(t)\cdot g}$
Here, $W$ is the Wronskian.

Take the case where $D\gt0$ . Then the characteristic equation two solutions, say $\lambda_1$ and $\lambda_2$ . These solutions satisfy the equations $\lambda_1+\lambda_2=-p$ and $\lambda_1\cdot \lambda_2=q$ . We can take as the basis solutions of the homogeneous equation: $y_1(t)=\e^{\lambda_1\cdot t}$ and $y_2(t)=\e^{\lambda_2\cdot t}$ . The Wronskian of this pair of differentiable functions is
$W(t)= (\lambda_2-\lambda_1)\cdot \e^{(\lambda_1+\lambda_2)\cdot t}= (\lambda_2-\lambda_1)\cdot \e^{-p\cdot t}$

We are ready to calculate the particular solution:
$\begin{array}{rcl} c_1'(t) &=&-\dfrac{1}{W(t)}\cdot y_2(t)\cdot g(t)\\ &=&-\dfrac{1}{\lambda_2-\lambda_1}\cdot\e^{p\cdot t}\cdot \e^{\lambda_2\cdot t}\cdot g\\ &&\phantom{x}\color{blue}{W(t) = (\lambda_2-\lambda_1)\cdot \e^{-p\cdot t}}\\ &=&-\dfrac{g}{\lambda_2-\lambda_1}\cdot\e^{-\lambda_1\cdot t}\\ &&\phantom{x}\color{blue}{\lambda_1+\lambda_2=-p}\\ &\text{and,}&\text{ likewise, }\\ c_2'(t)&=&\dfrac{1}{W(t)}\cdot y_1(t)\cdot g(t)\\&=&\dfrac{1}{\lambda_2-\lambda_1}\cdot\e^{p\cdot t}\cdot \e^{\lambda_1\cdot t}\cdot g\\&=&\dfrac{g}{\lambda_2-\lambda_1}\cdot\e^{-\lambda_2\cdot t}\end{array}$
Because $g$ is a constant, we find by integration,
$\begin{array}{rcl}c_1(t)&=&\displaystyle-\frac{g}{\lambda_2-\lambda_1}\cdot\int \e^{-\lambda_1\cdot t}\,\dd t=\frac{g}{\lambda_1\cdot(\lambda_2-\lambda_1)}\cdot\e^{-\lambda_1\cdot t}\\ &\text{and}&\\ c_2(t)&=&\displaystyle\frac{g}{\lambda_2-\lambda_1}\cdot\int \e^{-\lambda_2\cdot t}\,\dd t=\frac{g}{-\lambda_2\cdot(\lambda_2-\lambda_1)}\cdot\e^{-\lambda_2\cdot t}\end{array}$
The conclusion is
$\begin{array}{rcl}y(t)&=&c_1(t)\cdot y_1(t)+c_2(t)\cdot y_2(t)\\ &=&\displaystyle \frac{g}{\lambda_1\cdot(\lambda_2-\lambda_1)}\cdot\e^{-\lambda_1\cdot t}\cdot\e^{\lambda_1\cdot t} +\frac{g}{-\lambda_2\cdot(\lambda_2-\lambda_1)}\cdot\e^{-\lambda_2\cdot t} \cdot\e^{\lambda_2\cdot t} \\ &=&\dfrac{g}{\lambda_2-\lambda_1}\cdot \left(\dfrac{1}{\lambda_1} -\dfrac{1}{\lambda_2}\right) \\ &=&\dfrac{g}{\lambda_2-\lambda_1}\cdot \dfrac{\lambda_2-\lambda_1}{\lambda_1\cdot\lambda_2} \\ &=&\dfrac{g}{q}\\ &&\phantom{x}\color{blue}{\lambda_1\cdot \lambda_2=q} \end{array}$
Verification for the other two cases ( $D=0$ and $D\lt0$ ) goes along the same lines.

This solution can also be found by use of the first example of the Ansatz method for linear second-order GDVs.

New example