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17calculus > differential equations > variation of parameters

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Wikipedia - Variation of Parameters

Variation of Parameters

\(\displaystyle{ y'' + p(t)y' + q(t)y = g(t) }\)

classify - second-order, linear, inhomogeneous

Alternate Name For Variation of Parameters: Variation of Constants

Variation of Parameters is a second order technique to solve nonhomogeneous differential equations. The form of the equations we solve using this technique is
\( y'' + p(t)y' + q(t)y = g(t) \) where \( g(t) \neq 0 \).

Although this technique will work when \( g(t) \) is a standard form, we normally use the method of undetermined coefficients which is easier. However, variation of parameters can be used for a wider range of problems.
The equations we use are

\( W = \begin{vmatrix} y_1(t) & y_2(t) \\ y'_1(t) & y'_2(t) \end{vmatrix} \)

\(\displaystyle{ u'_1(t) = \frac{ \begin{vmatrix} 0 & y_2(t) \\ g(t) & y'_2(t) \end{vmatrix} }{W} }\)

\(\displaystyle{ u'_2(t) = \frac{ \begin{vmatrix} y_1(t) & 0 \\ y'_1(t) & g(t) \end{vmatrix} }{W} }\)

\(y_1(t)\) and \(y_2(t)\) are solutions to the homogeneous equation

The matrix \(W\) is called the Wronskian, and from the above equations you can tell that we require \(W \neq 0\). After evaluating the above matrices for \(u'_1(t)\) and \(u'_2(t)\), we integrate to obtain \(u_1(t)\) and \(u_2(t)\). The final solution to the differential equation is

\( y (t) = u_1(t) y_1 (t) + u_2(t) y_2 (t) \)

It will help you understand these if you know where the equations come from. So the next panel shows the derivation of the these equations and explains step-by-step why we can use them in this form. If you prefer, just below the panel is a video explaining this same derivation.

Deriving the Variation of Parameters Equations

1. Set Up Equations

First, we form the homogeneous ( \(g(t)=0\) ) solution
\( y_H (t) = c_1 y_1 (t) + c_2 y_2 (t) \).
However, we replace the constants \( c_1, c_2\) with functions \( u_1 (t), u_2 (t) \). This gives us

\( y (t) = u_1(t) y_1 (t) + u_2(t) y_2 (t) \)       [1]

Let's take the derivative with respect to \(t\).

\( y'(t) = u_1(t) y'_1(t) + \color{red}{u'_1(t)y_1(t)} + \) \( u_2(t) y'_2(t) + \color{red}{u'_2(t)y_2(t)} \)

Now we set up an unusual condition; we require

\( u'_1(t)y_1(t) + u'_2(t)y_2(t) = 0 \)

i.e. we take the \(u\)-derivative terms (red terms) and set the sum equal to zero. This removes those terms from the derivative \( y'(t) \) and we are left with
\( y'(t) = u_1(t) y'_1(t) + u_2(t) y'_2(t) \). Let's take the derivative of this again.

\( y''(t) = u_1(t) y''_1(t) + \color{red}{u'_1(t) y'_1(t)} + \) \( u_2(t) y''_2(t) + \color{red}{u'_2(t) y'_2(t)} \)

Now we set up another condition.

\( u'_1(t) y'_1(t) + u'_2(t) y'_2(t) = g(t) \)

Again, we take the \(u\)-derivatives terms (red terms) and set the sum equal to \( g(t) \) from the original differential equation.

Finally, we have the two equations

\( \begin{array}{lclcl} u'_1(t)y_1(t) & + & u'_2(t)y_2(t) & = & 0 \\ u'_1(t) y'_1(t) & + & u'_2(t) y'_2(t) & = & g(t) \end{array} \)

with the two unknowns \( u'_1(t), u'_2(t) \).

2. Solving For Two Unknowns

Using Cramer's Rule to solve a system of equations, we have the Wronskian

\( W = \begin{vmatrix} y_1(t) & y_2(t) \\ y'_1(t) & y'_2(t) \end{vmatrix} \)     which we use to solve for \( u'_1(t) \) and \( u'_2(t) \)


\(\displaystyle{ u'_1(t) = \frac{ \begin{vmatrix} 0 & y_2(t) \\ g(t) & y'_2(t) \end{vmatrix} }{W} }\)

   

\(\displaystyle{ u'_2(t) = \frac{ \begin{vmatrix} y_1(t) & 0 \\ y'_1(t) & g(t) \end{vmatrix} }{W} }\)

3. Finishing Up

Now we integrate the equations for \( u'_1(t)\) and \( u'_2(t) \) and plug the results into equation [1] to get our final answer. If we have initial conditions, we would need to use them to find the constants that result from the integration in the last step.

There are two ways to think about how to write the final solution. You can integrate \( u'_1(t)\) and \( u'_2(t) \) to get \( u_1(t) + c_1\) and \( u_2(t) + c_2 \) and write the final solution as
\( y (t) = (u_1(t) + c_1) y_1 (t) + \) \( ( u_2(t) + c_2) y_2 (t) \)
or, as the video below shows, you can think of the solution \( y_p (t) = u_1(t) y_1 (t) + u_2(t) y_2 (t) \) as a particular solution to which you add the homogeneous solution to get
\( y (t) = y_p(t) + y_h(t) = \) \( u_1(t) y_1 (t) + u_2(t) y_2 (t) + \) \( c_1y_1(t) + c_2y_2(t) \)
As you can see, these are just two different ways of looking at the same thing. It is best to follow what your instructor requires.

Here is a video showing the same derivation. The presenter does a very good job explaining the equations.

Dr Chris Tisdell - variation of parameters

Here is another video discussing variation of parameters. This video is quite in-depth and he starts out with some matrix theory that supports the later discussion. This video is included here for completeness but is not necessary for using variation of parameters.

MIT OCW - variation of parameters

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Practice Problems

Instructions - - Unless otherwise instructed, solve these differential equations using the variation of parameters technique. Give your answers in exact terms and completely factored.

Level A - Basic

Practice A01

\(y''-5y'+6y=4e^t\)

answer

solution

Practice A02

\(y_1=x\) and \(y_2=x^3\) are solutions to the homogeneous equation, find the general solution of \(x^2y''-3xy'+3y=4x^7\)

answer

solution


Level B - Intermediate

Practice B01

\(\displaystyle{y''-2y'+y=\frac{e^x}{x^2+1}}\)

answer

solution

Practice B02

\(y''+y=\tan(x)\)

answer

solution

Practice B03

\(y''+4y=\csc(2x)\)

answer

solution

Practice B04

\(\displaystyle{y''+y=\frac{1}{\cos(x)}}\)

answer

solution

Practice B05

\(\displaystyle{y''-2y'+y=\frac{e^x}{x^4}}\)

answer

solution


Level C - Advanced

Practice C01

\(y''+9y=2\tan(3x)\)

answer

solution

Practice C02

\(y''+3y'+2y=5\cos(t)\)

answer

solution

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