Variation of Parameters

Section 3.9 Variation of Parameters

Consider the equation

$\begin{equation*} y^{\prime\prime}+4y =\frac{3}{\sin t} \end{equation*}$

MOUC doesn't work with quotients, only products. We will learn a (more complicated) general formula to solving more general linear non-homoheneous 2nd order ODEs. This formula will give a particular solution to a non-homogeneous equation given that you already know what the fundamental set of solutions are for the corresponsing homogeneous equation.

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Theorem 3.9.1. Variation of Parameters.

If $p,q,$ and $g$ are continuous on an open interval $I\text{,}$ and if the functions $\left\{ y_{1},y_{2}\right\}$ form a fundamental set of solutions to the corresponding homogeneous equation

$\begin{equation*} y^{\prime\prime}+p(t)y^{\prime}+q(t)y=0, \end{equation*}$

then a particular solution to

$\begin{equation*} y^{\prime\prime}+p(t)y^{\prime}+q(t)y=g(t) \end{equation*}$

is given by

$\begin{align*} y_{p}(t) \amp =-y_{1}(t)\int_{t_{0}}^{t}\frac{y_{2}(s)g(s)}{W\left(y_{1},y_{2}\right)(s)}ds+y_{2}(t)\int_{t_{0}}^{t}\frac{y_{1}(s)g(s)}{W\left(y_{1},y_{2}\right)(s)}ds\\ \amp =-y_{1}(t)\left[\int\frac{y_{2}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]+y_{2}(t)\left[\int\frac{y_{1}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right] \end{align*}$

if the antiderivates exist, where $t_{0}$ is any value in $I\text{.}$ Then the general solution to the non-homogeneous solution is

$\begin{equation*} y(t)=c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t). \end{equation*}$

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Proof.

The proof can be found in any differential equations text, or online. The idea is this: Suppose

\begin{equation*} y_{h}(t)=c_{1}y_{1}(t)+c_{2}y_{1}(t) \end{equation*}

is the general solution to

\begin{equation*} y^{\prime\prime}+p(t)y^{\prime}+q(t)y=0. \end{equation*}

Then the idea is to use the following guess:

\begin{equation*} y_{p}(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t) \end{equation*}

for the non-homogeneous equation, and also make the extra assumption that

\begin{equation*} u_{1}^{\prime}(t)y_{1}(t)+u_{2}^{\prime}(t)y_{2}(t)=0.\,\,\,\,\,(\star) \end{equation*}

The validity of this assumption is difficult to justify without higher level mathematics, but one can at least take comfort in that we have an extra constraint to play with, so for computational convenience we select condition $(\star)\text{.}$

Then take derivatives, simplify and put them back into the differential equation. This will always reduce to

\begin{align*} LHS \amp =y_{p}^{\prime\prime}+p(t)y_{p}^{\prime}+q(t)y_{p}\\ \amp =\text{work}\\ \amp =u_{1}^{\prime}y_{1}^{\prime}(t)+u_{2}^{\prime}(t)y_{2}^{\prime}(t) \end{align*}

and set LHS to RHS which is $g(t)$ hence we get

\begin{equation*} u_{1}^{\prime}(t)y_{1}^{\prime}(t)+u_{2}^{\prime}(t)y_{2}^{\prime}(t)=g(t).\,\,\,\,\,(\star\star) \end{equation*}

Putting $(\star)$ and $(\star\star)$ together we have the two equations:

\begin{equation*} \begin{cases} u_{1}^{\prime}(t)y_{1}(t)+u_{2}^{\prime}(t)y_{2}(t)=0\\ u_{1}^{\prime}y_{1}^{\prime}(t)+u_{2}^{\prime}(t)y_{2}^{\prime}(t)=g(t) \end{cases} \end{equation*}

which boils to solving for $u_{1}^{\prime}(t)$ and $u_{2}^{\prime}(t)$ and getting

\begin{equation*} \begin{cases} u_{1}^{\prime}(t)=-\frac{y_{2}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}\\ u_{2}^{\prime}(t)=\frac{y_{1}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)} \end{cases} \end{equation*}

which by integrating we have

\begin{equation*} \begin{cases} u_{1}(t)=-\int\frac{y_{2}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\\ u_{2}(t)=\int\frac{y_{1}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt \end{cases} \end{equation*}

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Example 3.9.2.

Find a particular solution to

\begin{equation*} y^{\prime\prime}+4y=\frac{1}{\cos\left(2t\right)}. \end{equation*}

Step1: First find $y_{h}$ if possible. In this case $y_{h}$will be given by solving $r^{2}+4=0$ so that $r=\pm2i$ hence

\begin{equation*} y_{h}(t)=c_{1}\cos(2t)+c_{2}\sin\left(2t\right). \end{equation*}

Thus $y_{1}(t)=\cos(2t)$ and $y_{2}(t)=\sin\left(2t\right)\text{.}$

Step2: Find the Wronskian:

\begin{align*} W(y_{1},y_{2})(t) \amp =\left|\begin{array}{cc} \cos(2t) \amp \sin\left(2t\right)\\ -2\sin(2t) \amp 2\cos(2t) \end{array}\right|\\ \amp =2\cos^{2}(2t)+2\sin^{2}(2t)\\ \amp =2\left[\cos^{2}(2t)+\sin^{2}(2t)\right]\\ \amp =2\cdot1=2. \end{align*}

Step3: Use our formula with $g(t)=\frac{1}{\cos\left(2t\right)}$ and get

\begin{align*} y_{p}(t) \amp =-y_{1}(t)\left[\int\frac{y_{2}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]+y_{2}(t)\left[\int\frac{y_{1}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]\\ \amp =-\cos(2t)\left[\int\frac{1}{2}\frac{\sin(2t)}{\cos\left(2t\right)}dt\right]+\sin(2t)\left[\int\frac{\cos(2t)}{2}\frac{1}{\cos\left(2t\right)}dt\right]\\ \amp =-\cos(2t)\left[\frac{1}{2}\int\frac{\sin(2t)}{\cos\left(2t\right)}dt\right]+\frac{t}{2}\sin(2t) \end{align*}

Now you can remember the antiderivative of $\int\tan(2t)dt$ or use $u$-substitution with $u=\cos(2t)$ and get $du=-2\sin(2t)dt$ so that

\begin{equation*} \int\frac{\sin(2t)}{\cos\left(2t\right)}dt=-\frac{1}{2}\int\frac{du}{u}=-\frac{1}{2}\ln\left|u\right|=-\frac{1}{2}\ln\left|\cos(2t)\right| \end{equation*}

hence

\begin{equation*} y_{p}(t)=\frac{1}{4}\cos(2t)\ln\left|\cos(2t)\right|+\frac{t}{2}\sin(2t). \end{equation*}

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Example 3.9.3.

Find the general solution to

\begin{equation*} t^{2}y^{\prime\prime}+2ty^{\prime}-2y=6t \end{equation*}

given that

\begin{equation*} y_{1}(t)=t,\,\,\,y_{2}(t)=t^{-2} \end{equation*}

forms a fundamental set of solution for the corresponding homogeneous differential equation.

Step 1: Since $y_{1}(t)=t,\,\,\,y_{2}(t)=t^{-2}$ forms a fundamental set of solution, this means that the general solution for the homogeneous equation is

\begin{equation*} y_{h}=c_{1}t+c_{2}t^{-2}. \end{equation*}

Step 2: Find the Wronskian:

\begin{align*} W(y_{1},y_{2})(t) \amp =\left|\begin{array}{cc} t \amp t^{-2}\\ 1 \amp -2t^{-3} \end{array}\right|\\ \amp =-2t^{-2}-t^{-2}=-3t^{-2}\neq0, \end{align*}

Step 3: Rewrite the equation in the form $y^{\prime\prime}+p(t)y^{\prime}+q(t)y=g(t)$ and hence

\begin{equation*} y^{\prime\prime}+\frac{2}{t}y^{\prime}-\frac{2}{t^{2}}y=\frac{6}{t}. \end{equation*}

Use our formula with $g(t)=\frac{6}{t}$ and get

\begin{align*} y_{p}(t) \amp =-y_{1}(t)\left[\int\frac{y_{2}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]+y_{2}(t)\left[\int\frac{y_{1}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]\\ \amp =-t\left[\int\frac{t^{-2}}{-3t^{-2}}\frac{6}{t}dt\right]+t^{-2}\left[\int\frac{t}{-3t^{-2})}\frac{6}{t}dt\right]\\ \amp =-t\left[\int\frac{2}{t}dt\right]+t^{-2}\left[\int-2t^{2}dt\right]\\ \amp =-t\left[2\ln t\right]+t^{-2}\left[-\frac{2}{3}t^{3}\right]\\ \amp =-2t\ln t-\frac{2}{3}t. \end{align*}

Hence, the general solution is

\begin{align*} y(t) \amp =y_{h}+y_{p}\\ \amp =c_{1}t+c_{2}t^{-2}-2t\ln t-\frac{2}{3}t. \end{align*}

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Example 3.9.4.

Find the general solution to

\begin{equation*} t^{2}y^{\prime\prime}-3ty^{\prime}+3y=8t^{3},\,\,\,\,t>0 \end{equation*}

given that

\begin{equation*} y_{1}(t)=t,\,\,\,y_{2}(t)=t^{3} \end{equation*}

forms a fundamental set of solution for the corresponding homogeneous differential equation.

Step 1: Since $y_{1}(t)=t,\,\,\,y_{2}(t)=t^{3}$ forms a fundamental set of solution, this means that the general solution for the homogeneous equation is

\begin{equation*} y_{h}=c_{1}t+c_{2}t^{3}. \end{equation*}

Step 2: Find the Wronskian:

\begin{align*} W(y_{1},y_{2})(t) \amp =\left|\begin{array}{cc} t \amp t^{3}\\ 1 \amp 3t^{2} \end{array}\right|\\ \amp =3t^{3}-t^{3}=2t^{3}\neq0, \end{align*}

Step 3: Rewrite the equation in the form $y^{\prime\prime}+p(t)y^{\prime}+q(t)y=g(t)$ and hence

\begin{equation*} y^{\prime\prime}-\frac{3}{t}y^{\prime}+\frac{3}{t^{2}}y=8t,. \end{equation*}

Use our formula with $g(t)=8t$ and get

\begin{align*} y_{p}(t) \amp =-y_{1}(t)\left[\int\frac{y_{2}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]+y_{2}(t)\left[\int\frac{y_{1}(t)g(t)}{W\left(y_{1},y_{2}\right)(t)}dt\right]\\ \amp =-t\left[\int\frac{t^{3}}{2t^{3}}8tdt\right]+t^{3}\left[\int\frac{t}{2t^{3}}8tdt\right]\\ \amp =-t\left[\int4tdt\right]+t^{3}\left[\int\frac{4}{t}dt\right]\\ \amp =-t\left[2t^{2}\right]+t^{3}\left[4\ln t\right]\\ \amp =-2t^{3}+4t^{3}\ln t \end{align*}

hence the general solution is

\begin{align*} y(t) \amp =y_{h}+y_{p}\\ \amp =c_{1}t+c_{2}t^{3}-2t^{3}+4t^{3}\ln t. \end{align*}