Math292 Textbook Variation of Parameters

Section 1.1 Variation of Parameters

Straight to the point....

Variation of Parameters tells us that the ODE $p (t) y^{″} + q (t) y^{'} + r (t) y = g (t)$ has a particular solution $Y (t) = u_{1} (t) y_{1} (t) + u_{2} (t) y_{2} (t),$ where

\begin{matrix} Y (t) = \int_{t_{0}}^{t} \frac{y_{1} (s) y_{2} (t) - y_{1} (t) y_{2} (s)}{W [y_{1}, y_{2}] (s)} \cdot \frac{g (s)}{p (s)} d s . \end{matrix}

Here, we take ${y_{1}, y_{2}}$ to be a fundamental set of solutions of the homogeneous ODE $p (t) y^{″} + q (t) y^{'} + r (t) y = 0,$ that is to say, $y_{1} y_{2}^{'} - y_{1}^{'} y_{2} \neq 0 .$

We also take $t_{0}$ to be some number that lies in the range of $t$ that the ODE is defined on. Choosing an appropriate $t_{0}$ can sidestep a lot of the computational work.

Another way one can look at Variation of Parameters is that, given $Y (t) = u_{1} (t) y_{1} (t) + u_{2} (t) y_{2} (t),$ the functions $u_{1}$ and $u_{2}$ satisfy the equation

\begin{array}{r} [\begin{array}{c} y_{1} & y_{2} \\ y_{1}^{'} & y_{2}^{'} \end{array}] [\begin{array}{c} u_{1}^{'} \\ u_{2}^{'} \end{array}] = [\begin{array}{c} 0 \\ g (t) / p (t) \end{array}] . \end{array}

Memorizing this (or the process below) is certainly less meticulous than memorizing the direct integral form above. Notice that the $2 \times 2$ matrix on the left hand side above is the matrix whose determinant equals the $W [y_{1}, y_{2}] (t) .$

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We are often interested in finding a particular solution of a nonhomogenous linear ODE

p (t) y^{″} + q (t) y^{'} + r (t) y = g (t),

mainly because the general solution of this ODE is

\begin{matrix} y (t) = h (t) + Y (t), \end{matrix}

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where

h (t)

is the general solution of the homogeneous linear ODE

p (t) y^{″} + q (t) y^{'} + r (t) y = 0

and

Y (t)

is any particular solution of the nonhomogeneous ODE

p (t) y^{″} + q (t) y^{'} + r (t) y = g (t) .

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Using Variation of Parameters or Duhamel's Principle helps us find some particular solution

Y (t)

for the nonhomogeneous ODE. Here, we'll start with Variation of Parameters.

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Subsection 1.1.1 Starting Variation of Parameters

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Variation of Parameters begins by assuming that a particular solution

Y (t)

\begin{matrix} p (t) y^{″} + q (t) y^{'} + r (t) y = g (t) \end{matrix}

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is of the form

\begin{matrix} Y (t) = u_{1} (t) y_{1} (t) + u_{2} (t) y_{2} (t) . \end{matrix}

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Here,

y_{1}

and

y_{2}

are fundamental solutions to the homogeneous ODE

p (t) y^{″} + q (t) y^{'} + r (t) y = 0,

that is, the Wronskian

W [y_{1}, y_{2}] (t) \neq 0

for all applicable

t .

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Review of Wronskian.

Let $y_1, y_2$ be solutions of the homogeneous ODE $p(t) y'' + q(t) y' + r(t)y = 0\text{.}$ Recall that the Wronskian of $y_1, y_2$ is denoted $W[y_1, y_2](t)\text{,}$ and

\begin{align*} W[y_1, y_2](t) = \det \begin{bmatrix} y_1(t) \amp y_2(t) \\ y_1'(t) \amp y_2'(t) \end{bmatrix} = y_1(t) y_2'(t) - y_1'(t) y_2(t). \end{align*}

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The idea behind Variation of Parameters is to kind of "generalize" the general solution of the homogenous ODE. Note that if we have that

u_{1} (t) = c_{1}

and

u_{2} (t) = c_{2}

are constant for all

t,

then the solution

Y (t) = c_{1} y_{1} (t) + c_{2} y_{2} (t)

is actually a solution of the homogeneous ODE!

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So, we generalize the homogeneous solution by

Y (t) = u_{1} (t) y_{1} (t) + u_{2} (t) y_{2} (t)

in hopes that we find one (we only need one particular nonhomogeneous solution to describe them all!) solution to the nonhomogeneous ODE.

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The algebra here can get a bit tricky, but I'll do my best to describe all the steps used!

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Subsection 1.1.2 Equations involving $u_{1} (t)$ and $u_{2} (t)$

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Now, since

Y (t)

is a particular solution of

p (t) y^{″} + q (t) y^{'} + r (t) y = g (t),

we have that

\begin{matrix} (1.1.1) & p (t) Y^{″} (t) + q (t) Y^{'} (t) + r (t) Y (t) = g (t) . \end{matrix}

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We'll want to calculate

Y^{'} (t)

and

Y^{″} (t),

using the form

Y (t) = u_{1} (t) y_{1} (t) + u_{2} (t) y_{2} (t) .

Doing so will create equations for

u_{1}

and

u_{2},

which we will hopefully be able to solve.

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First, calculate

Y^{'} (t) :

(try it yourself! Remember that

Y (t) = u_{1} y_{1} + u_{2} y_{2},

all of those terms being functions of

t .

)

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Question 1.1.1. What's $Y^{'} (t)$ equal to?

We make good use of the product rule! Note that

\begin{align*} Y'(t) \amp = \Big[ u_1'(t) y_1(t) + u_2'(t) y_2(t) \Big] + u_1(t) y_1'(t) + u_2(t) y_2'(t). \end{align*}

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We have 4 terms in our representation of

Y^{'} (t) .

This means that

Y^{″} (t)

will have 8 terms! To reduce down on the number of terms we have, we set

\begin{matrix} u_{1}^{'} (t) y_{1} (t) + u_{2}^{'} (t) y_{2} (t) = 0. \end{matrix}

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This may seem arbitrary for now, but this will ease up a lot of our calculations. We now have that

\begin{aligned} Y^{'} (t) & = [u_{1}^{'} (t) y_{1} (t) + u_{2}^{'} (t) y_{2} (t)] + u_{1} (t) y_{1}^{'} (t) + u_{2} (t) y_{2}^{'} (t) \\ = u_{1} (t) y_{1}^{'} (t) + u_{2} (t) y_{2}^{'} (t) . \end{aligned}

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Now, we calculate

Y^{″} (t),

using our representation of

Y^{'} (t)

above.

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Question 1.1.2. What's $Y^{″} (t)$ equal to?

Again, this should be another quick differentiation exercise.

\begin{gather*} Y''(t) = \Big[ u_1'(t) y_1'(t) + u_2'(t) y_2'(t) \Big] + u_1(t) y_1''(t) + u_2(t) y_2''(t). \end{gather*}

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Now, we have expressions for

Y^{'} (t)

and

Y^{″} (t) .

Substitute these into (1.1.1) (Why isn't this working? This should expand out to show equation (1.1.1), will fix later). Doing so would yield that

\begin{matrix} u_{1}^{'} (t) y_{1}^{'} (t) + u_{2}^{'} (t) y_{2}^{'} (t) = \frac{g (t)}{p (t)} . \end{matrix}

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Question 1.1.3. Why does this occur?

Since $y_1$ and $y_2$ are solutions of the homogeneous equation, notice that

\begin{gather*} p(t) y_1''(t) + q(t) y_1'(t) + r(t )y_1(t) = 0, \end{gather*}

and

\begin{gather*} p(t) y_2''(t) + q(t) y_2'(t) + r(t )y_2(t) = 0. \end{gather*}

Now, we directly substitute $Y'(t)$ and $Y''(t)$ into (1.1.1):

\begin{align*} g(t) \amp = p(t) \Big[ u_1'(t) y_1'(t) + u_2'(t) y_2'(t) + u_1(t) y_1''(t) + u_2(t) y_2''(t) \Big] \\ \amp \quad + \ q(t) \Big[ u_1(t) y_1'(t) + u_2(t) y_2'(t) \Big] \\ \amp \quad + \ r(t) \Big[ u_1(t) y_1(t) + u_2(t) y_2(t) \Big], \\ \amp = u_1(t)\Big[ p(t) y_1''(t) + q(t) y_1'(t) + r(t )y_1(t) \Big] \\ \amp \quad + \ u_2(t) \Big[ p(t) y_2''(t) + q(t) y_2'(t) + r(t )y_2(t) \Big] \\ \amp \quad + \ p(t) \Big[ u_1'(t) y_1'(t) + u_2'(t) y_2'(t) \Big], \\ \amp = p(t) \Big[ u_1'(t) y_1'(t) + u_2'(t) y_2'(t) \Big]. \end{align*}

Finish off by dividing $p(t)$ on both sides. Note that we must have that $p(t) \neq 0$ for all $t$ in the appropriate interval.

With this, we have finally found two equations involving

u_{1}

and

u_{2},

which we can solve as a typical system of linear equations:

\begin{aligned} u_{1}^{'} y_{1} + u_{2}^{'} y_{2} & = 0, \\ u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} & = \frac{g (t)}{p (t)} . \end{aligned}

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Subsection 1.1.3 Solving for $u_{1} (t)$ and $u_{2} (t)$

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We can write the two equations from the previous section as a matrix system.

\begin{array}{r} [\begin{array}{c} y_{1} & y_{2} \\ y_{1}^{'} & y_{2}^{'} \end{array}] [\begin{array}{c} u_{1}^{'} \\ u_{2}^{'} \end{array}] = [\begin{array}{c} 0 \\ g (t) / p (t) \end{array}] \end{array}

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With our knowledge from linear algebra, we can solve for

u_{1}^{'}

and

u_{2}^{'}

using matrix inverses:

\begin{aligned} [\begin{array}{c} u_{1}^{'} \\ u_{2}^{'} \end{array}] & = {[\begin{array}{c} y_{1} & y_{2} \\ y_{1}^{'} & y_{2}^{'} \end{array}]}^{- 1} [\begin{array}{c} 0 \\ g (t) / p (t) \end{array}] \\ = \frac{1}{y_{1} y_{2}^{'} - y_{1}^{'} y_{2}} [\begin{array}{c} y_{2}^{'} & - y_{2} \\ - y_{1}^{'} & y_{1} \end{array}] [\begin{array}{c} 0 \\ g (t) / p (t) \end{array}] \\ = \frac{g (t) / p (t)}{W [y_{1}, y_{2}] (t)} [\begin{array}{c} - y_{2} \\ y_{1} \end{array}] \end{aligned}

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At last! We have solved for

u_{1}^{'} (t)

and

u_{2}^{'} (t) :

\begin{matrix} u_{1}^{'} (t) = - \frac{y_{2} (t)}{W [y_{1}, y_{2}] (t)} \frac{g (t)}{p (t)} and u_{2}^{'} (t) = \frac{y_{1} (t)}{W [y_{1}, y_{2}] (t)} \frac{g (t)}{p (t)} . \end{matrix}

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We want

u_{1}

and

u_{2} .

We can choose

t_{0}

to be some number for which

u_{1} (t_{0}) = 0

and

u_{2} (t_{0}) = 0 .

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Question 1.1.4. Why can we choose $t_{0}$ to satisfy $u_{1} (t_{0}) = u_{2} (t_{0}) = 0 ?$

When we perform "indefinite integrals" we always have to add a constant to our answer (that "$+ C$"). For example, we would write

\begin{gather*} \int x^2 \ dx = \frac{x^3}3 + C. \end{gather*}

In general, we have that

\begin{gather*} \int f'(x) \ dx = f(x) + C. \end{gather*}

That is to say, given $f'(x)\text{,}$ we can find infinitely many unique $f(x)\text{.}$ As in the case before, any function of the form $x^3/3 + C\text{,}$ for any constant $C\text{,}$ has the derivative $x^2$

All this is to say that we know what $u_1'$ and $u_2'$ are. But, we can find infinitely many unique $u_1$ and $u_2$ whose derivatives satisfy the forms we have. So we can shift $u_1$ and $u_2$ up or down until both intersect at some $t_0$ with a value of $0\text{.}$

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This helps us in that we can write that