Theory

Introduction

The method of variation of parameters is to change a constant value parameter like $C_1$ into a variable function $v_1(t)$, and see whether we can cook up an appropriate $v_1(t)$ so that the solution is still a solution with this variable parameter $v_1(t)$.

Discovering a second solution

Suppose we are given a single solution $y_1(t)$ to the homogeneous equation $y^{\prime \prime }+p{y}^{\prime }+qy=0$. We know that $c_1{y}_1(t)$ is also a solution, but it isn’t independent of $y_1(t)$. We would like to find another, independent, solution $y_2(t)$ which will span (along with $y_1(t)$) the complete set of solutions.

Let us propose the hypothesis that there is a solution $y_2(t)=v(t)y_1(t)$ for $v(t)$ some function of $t$.

This solution $y_2(t)$ will be independent of $y_1(t)$ provided $v(t)$ is not constant. The trick is to supply some additional constraints on $v(t)$ that make it feasible to solve for what $v(t)$ must be, given that $v(t)y_1(t)$ solves the equation. These constraints emerge from the process of plugging $v(t)y_1(t)$ into the equation and expanding the product and chain rules.

$$\begin{array}{rl}(v{y}_1{)}^{\prime \prime }+p(v{y}_1{)}^{\prime }+q\left(v{y}_1\right)& =0\\ v^{\prime \prime }{y}_1+2v^{\prime }{y}_1^{\prime }+v{y}_1^{\prime \prime }+p(v^{\prime }{y}_1+v{y}_1^{\prime })+qv{y}_1& =0\\ v(y_1^{\prime \prime }+p{y}_1^{\prime }+q{y}_1)+v^{\prime \prime }{y}_1+v^{\prime }(2y_1^{\prime }+p{y}_1)& =0,\end{array}$$ $$\frac{v^{\prime \prime }}{v^{\prime }}=-2\frac{y_1^{\prime }}{y_1}-p,$$

which we continue to solve as by separation of variables:

$$\begin{array}{rl}\ln v^{\prime }& =-2\ln y_1-\int pdt\\ v^{\prime }& =y_1^{-2}{e}^{-\int pdt}\\ v(t)& =\int y_1^{-2}{e}^{-\int pdt}dt.\end{array}$$

(We have allowed an abuse of notation by letting $\int pdt$ indicate any particular antiderivative, understood to be a function of $t$.) By running these equations backwards, we find that if $v(t)=\int y_1^{-2}{e}^{-\int pdt}dt$, then $y_2=v{y}_1$ is indeed a second solution.

This method always works! That is, it provides a second solution in terms of a first solution and two integrations.

This method is enough to explain the $t$ factor appearing in our secondary solutions $t{e}^{rt}$ to linear equations with constant coefficients and repeated roots:

Particular solution of inhomogeneous

The same strategy, with a complication, will provide a particular solution to an inhomogeneous equation by variation of the parameters of the complete set of solutions to its corresponding homogeneous equation.

Suppose $c_1{y}_1(t)+c_2{y}_2(t)$ is the complete family of solutions to $y^{\prime \prime }+p{y}^{\prime }+qy=0$. We seek a particular solution to $y^{\prime \prime }+p{y}^{\prime }+qy=f$. Let us propose the hypothesis that a particular solution has the form

$$y_0(t)=v_1(t)y_1(t)+v_2(t)y_2(t).$$

As before, we substitute this form for $y_0$ into the equation and eliminate the consequences of $y_1$ and $y_2$ being solutions. We compute:

$$\begin{array}{rl}{y}_0& =v_1{y}_1+v_2{y}_2\\ y_0^{\prime }& =(v_1^{\prime }{y}_1+v_2^{\prime }{y}_2)+(v_1{y}_1^{\prime }+v_2{y}_2^{\prime })\\ y_0^{\prime \prime }& =(v_1^{\prime }{y}_1+v_2^{\prime }{y}_2{)}^{\prime }+v_1^{\prime }{y}_1^{\prime }+v_1{y}_1^{\prime \prime }+v_2^{\prime }{y}_2^{\prime }+v_2{y}_2^{\prime \prime }.\end{array}$$

Now for the new complication of this method. It will be very convenient to add the additional constraint that $v_1^{\prime }{y}_1+v_2^{\prime }{y}_2=0$. This is a constraint on the possibilities we are hypothetically contemplating for $y_0(t)$, meaning for $v_1$ and $v_2$. We are not required to add this constraint, but we will discover that it is not too strong in the sense that we can always succeed in finding a solution with this constraint, and it greatly simplifies the algebra, so we do so. The first terms in brackets in the expressions for $y_0^{\prime }$ and $y_0^{\prime \prime }$ therefore disappear.

Plugging the rest back into the equation, we find:

$$\begin{array}{rl}{y}_0^{\prime \prime }+p{y}_0^{\prime }+qy& =f(t)\\ v_1^{\prime }{y}_1^{\prime }+v_1{y}_1^{\prime \prime }+v_2^{\prime }{y}_2^{\prime }+v_2{y}_2^{\prime \prime }+p(v_1{y}_1^{\prime }+v_2{y}_2^{\prime })+q(v_1{y}_1+v_2{y}_2)& =f(t)\\ v_1(y_1^{\prime \prime }+p{y}_1^{\prime }+q{y}_1)+v_2(y_2^{\prime \prime }+p{y}_2^{\prime }+q{y}_2)+v_1^{\prime }{y}_1^{\prime }+v_2^{\prime }{y}_2^{\prime }& =f(t)\\ v_1^{\prime }{y}_1^{\prime }+v_2^{\prime }{y}_2^{\prime }& =f(t).\end{array}$$

So it will be enough to solve this last equation simultaneously with our constraint equation, namely to solve the system:

$$\{\begin{array}{rl}{v}_1^{\prime }{y}_1+v_2^{\prime }{y}_2& =0\\ v_1^{\prime }{y}_1^{\prime }+v_2^{\prime }{y}_2^{\prime }& =f.\end{array}$$

Given $y_1,y_2,f$, this is a simple system of linear equations in two unknowns $v_1^{\prime },v_2^{\prime }$ and we can solve for them explicitly:

$$\begin{array}{rl}{v}_1^{\prime }& =\frac{y_{2}f}{y_1^{\prime }{y}_2-y_1{y}_2^{\prime }},\\ v_2^{\prime }& =-\frac{y_{1}f}{y_1^{\prime }{y}_2-y_1{y}_2^{\prime }}.\end{array}$$

Therefore, setting

$$\begin{array}{rl}{v}_1& =\int \frac{y_{2}f}{y_1^{\prime }{y}_2-y_1{y}_2^{\prime }}dt,\\ v_2& =\int -\frac{y_{1}f}{y_1^{\prime }{y}_2-y_1{y}_2^{\prime }}dt,\end{array}$$

we have a solution $y_0=v_1{y}_1+v_2{y}_2$.