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Wikipedia - Variation of Parameters

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17calculus > differential equations > variation of parameters

General Form: $$y'' + p(t)y' + q(t)y = g(t)$$

Classification: 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 section shows the derivation of the these equations and explains step-by-step why we can use them in this form.

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 [13mins-40secs]

video by Dr Chris Tisdell

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.

video by MIT OCW

### Practice

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on.

GOT IT. THANKS!

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

Basic Problems

Use variation of parameters to find the general solution of $$y'' - 4y' + 4y = \sec^2(x)e^{2x}$$

Problem Statement

Use variation of parameters to find the general solution of $$y'' - 4y' + 4y = \sec^2(x)e^{2x}$$

$$y = [ -\ln|\cos(x)| + c_2 x + c_1 ] e^{2x}$$

Problem Statement

Use variation of parameters to find the general solution of $$y'' - 4y' + 4y = \sec^2(x)e^{2x}$$

Solution

Homogeneous solution first. $$r^2 - 4r + 4 = (r-2)^2 = 0 ~~~ \to ~~~ r = +2$$ multiplicity $$2$$

$$y_h = c_1 e^{2x} + c_2 x e^{2x}$$

Now, find the Wronskian.
$$\displaystyle{ W = \begin{vmatrix} y_{h1} & y_{h2} \\ y'_{h1} & y'_{h2} \end{vmatrix} = }$$ $$\displaystyle{ \begin{vmatrix} e^{2x} & xe^{2x} \\ 2e^{2x} & (2x+1)e^{2x} \end{vmatrix} = }$$ $$\displaystyle{ (2x+1)e^{4x} - 2xe^{4x} = e^{4x} }$$

Using variation of parameters, our non-homogeneous solution looks like $$y=u_1(x)e^{2x} + u_2(x)xe^{2x}$$.
Setting up the equations for $$u'_1(x)$$ and $$u'_2(x)$$, we have

$$\displaystyle{ u'_1 = \frac{ \begin{vmatrix} 0 & xe^{2x} \\ \sec^2 xe^{2x} & (2x+1)e^{2x} \end{vmatrix} }{W} = \frac{-x\sec^2 x e^{4x}}{e^{4x}} = -x\sec^2 x }$$

$$\displaystyle{ u'_2 = \frac{ \begin{vmatrix} e^{2x} & 0 \\ 2e^{2x} & \sec^2xe^{2x} \end{vmatrix} }{W} = \frac{\sec^2xe^{4x}}{e^{4x}} = \sec^2 x }$$

So we need to solve the two equations
$$\displaystyle{ u'_1 = -x \sec^2 x ~~~ \to ~~~ u_1 = -\int{x\sec^2 x ~ dx} }$$

$$\displaystyle{ u'_2 = \sec^2 x ~~~ \to ~~~ u_2 = \int{\sec^2 x~dx} }$$

The first integral is solved in the practice problems on the integration by parts page. So we just give the result here.
$$\displaystyle{ u_1 = -\int{x\sec^2 x ~ dx} = -x\tan(x)-\ln|\cos(x)|+k_1 }$$

The second one is easy, since $$(\tan(x))' = \sec^2(x)$$

$$\displaystyle{ u_2 = \int{\sec^2 x~dx} = \tan(x)+k_2 }$$

To get our final answer, we substitute these results into the equation $$y=u_1e^{2x} + u_2xe^{2x}$$ and simplify.
We also let $$c_1 = A+k_1$$ and $$c_2 = B+k_2$$ to combine the constants.

$$y = [ -\ln|\cos(x)| + c_2 x + c_1 ] e^{2x}$$

$$y'' - 5y' + 6y = 4e^t$$

Problem Statement

$$y'' - 5y' + 6y = 4e^t$$

$$y(t) = 2e^t + c_1e^{2t} + c_2e^{3t}$$

Problem Statement

$$y'' - 5y' + 6y = 4e^t$$

Solution

We could use the method of undetermined coefficients to solve this problem more easily. However, we are asked to use variation of parameters. We will work through the solution by going through the derivation of the equations.

 First, we solve the homogeneous equation. $$y'' - 5y' + 6y = 0$$ The characteristic equation is $$r^2 - 5r + 6 = 0$$ $$(r-3)(r-2) = 0$$ So our homogeneous solution is $$y_H(t) = c_1e^{2t} + c_2e^{3t}$$

Now, we work on the particular solution by replacing the constants with functions of t, $$y(t) = u_1(t)e^{2t} + u_2(t)e^{3t}$$, and taking the derivative with respect to t.
$$y'(t) = u_1(t)(2e^{2t}) + \color{red}{u'_1(t)e^{2t}} + u_2(t)(3e^{3t}) + \color{red}{u'_2(t)e^{3t} }$$
For our first equation, we take the terms in red and set them equal to zero to get $$u'_1(t)e^{2t} + u'_2(t)e^{3t} = 0$$     [ 1 ]
This leaves us with $$y'(t) = u_1(t)(2e^{2t}) + u_2(t)(3e^{3t})$$, which we now take the derivative of with respect to t giving us
$$y''(t) = 2u_1(t)(2e^{2t}) + \color{red}{ 2u'_1(t)e^{2t} } + 3u_2(t)(3e^{3t}) + \color{red}{ 3u'_2(t)e^{3t} }$$
Setting the red terms equal to $$g(t) = 4e^t$$, we have
$$2u'_1(t)e^{2t} + 3u'_2(t)e^{3t} = 4e^t$$
Now, we have the two equations we need, the last one and equation [1].

 $$u'_1(t)e^{2t} + u'_2(t)e^{3t} = 0$$ $$u'_1(t)(2e^{2t}) + u'_2(t)(3e^{3t}) = 4e^t$$

Now we need to solve for $$u'_1(t)$$ and $$u'_2(t)$$.

 $$\displaystyle{ W = \begin{vmatrix} e^{2t} & e^{3t} \\ 2e^{2t} & 3e^{3t} \end{vmatrix} }$$ $$W = 3e^{5t} - 2e^{5t} = e^{5t}$$ $$\displaystyle{ u'_1(t) = \frac{ \begin{vmatrix} 0 & e^{3t} \\ 4e^t & 3e^{3t} \end{vmatrix} }{W} }$$ $$\displaystyle{ u'_1(t) = \frac{0 -4e^{4t}}{e^{5t}} = -4e^{-t} }$$ $$\displaystyle{ u_1(t) = \int{-4e^{-t} dt} = 4e^{-t} + c_1 }$$ $$\displaystyle{ u'_2(t) = \frac{ \begin{vmatrix} e^{2t} & 0 \\ 2e^{2t} & 4e^t \end{vmatrix} }{W} }$$ $$\displaystyle{ u'_2(t) = \frac{4e^{3t}-0}{e^{5t}} = 4e^{-2t} }$$ $$u_2(t) = \int{4e^{-2t} dt} = -2e^{-2t} + c_2$$ Substituting these results into $$y(t) = u_1(t)e^{2t} + u_2e^{3t}$$ gives us $$y(t) = [ 4e^{-t} + c_1 ] e^{2t} + [ -2e^{-2t} + c_2 ] e^{3t}$$ $$y(t) = 4e^t+ c_1 e^{2t} - 2e^t + c_2 e^{3t}$$ $$y(t) = 2e^t + c_1 e^{2t} + c_2 e^{3t}$$

$$y(t) = 2e^t + c_1e^{2t} + c_2e^{3t}$$

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

Problem Statement

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

$$y = x^7 / 6 + c_1x + c_2x^3$$

Problem Statement

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

Solution

### 606 solution video

video by PatrickJMT

$$y = x^7 / 6 + c_1x + c_2x^3$$

$$2y'' - y' - y = 2e^t$$

Problem Statement

$$2y'' - y' - y = 2e^t$$

Solution

### 2218 solution video

Intermediate Problems

Find the general solution and use the method of variation of parameters to find the particular solution for $$y^{(3)} - 3y''+3y' - y = 6e^t$$.

Problem Statement

Find the general solution and use the method of variation of parameters to find the particular solution for $$y^{(3)} - 3y''+3y' - y = 6e^t$$.

$$y = Ae^t + Bte^t + Ct^2e^t + t^3e^t$$

Problem Statement

Find the general solution and use the method of variation of parameters to find the particular solution for $$y^{(3)} - 3y''+3y' - y = 6e^t$$.

Solution

This is a rather long problem but the 3x3 determinants reduce to manageable equations. We will show some intermediate results without all the details for calculating determinants.
For the homogeneous solution, we have constant coefficents, so our characteristic equation is $$r^3-3r^2+3r-1=0$$ $$\to$$ $$(r-1)^3=0$$. This gives us the homogeneous solution $$y_h=Ae^t+Bte^t+Ct^2e^t$$.
Now we need to set up the determinants for the variation of parameters method.

 $$y_1=e^t$$ $$y_2=te^t$$ $$y_3=t^2e^t$$

The Wronskian is
$$W = e^{3t} \begin{vmatrix} 1 & t & t^2 \\ 1 & t+1 & t^2+2t \\ 1 & t+2 & t^2+4t+2 \end{vmatrix}$$
This looks messy but it reduces to $$W=2e^{3t}$$. The variation of parameters technique gives us the following results.

 $$u'_1=3t^2 \to u_1=t^3$$ $$u'_2=-6t \to u_2=-3t^2$$ $$u'_3=3 \to u_3=3t$$

To get the particular solution, we use these results in the equation $$y_p = u_1 y_1 + u_2 y_2 + u_3 y_3 = t^3e^t$$ and our answer is $$y = y_h + y_p$$.

$$y = Ae^t + Bte^t + Ct^2e^t + t^3e^t$$

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

Problem Statement

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

$$\displaystyle{ y = Ae^x + Bxe^x - \frac{e^x}{2}\ln(x^2+1) + xe^x\arctan(x) }$$

Problem Statement

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

Solution

### 600 solution video

video by Dr Chris Tisdell

$$\displaystyle{ y = Ae^x + Bxe^x - \frac{e^x}{2}\ln(x^2+1) + xe^x\arctan(x) }$$

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

Problem Statement

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

$$y = -\cos(x)\ln\abs{\sec(x) + \tan(x)} + A\cos(x) + B\sin(x)$$

Problem Statement

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

Solution

Here are two video solutions by too different instructors. If, after watching the first video, you feel like you need more explanation, perhaps the second one will help.

### 601 solution video

video by Dr Chris Tisdell

### 601 solution video

$$y = -\cos(x)\ln\abs{\sec(x) + \tan(x)} + A\cos(x) + B\sin(x)$$

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

Problem Statement

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

$$\displaystyle{ y = A\cos(2x) + B\sin(2x) - \frac{x}{2}\cos(2x) + \frac{1}{4} \sin(2x) \ln\abs{\sin(2x)} }$$

Problem Statement

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

Solution

### 603 solution video

video by Dr Chris Tisdell

$$\displaystyle{ y = A\cos(2x) + B\sin(2x) - \frac{x}{2}\cos(2x) + \frac{1}{4} \sin(2x) \ln\abs{\sin(2x)} }$$

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

Problem Statement

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

$$y = A\cos(x) + B\sin(x) + \cos(x)\ln\abs{\cos(x)} + x\sin(x)$$

Problem Statement

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

Solution

### 604 solution video

video by Dr Chris Tisdell

$$y = A\cos(x) + B\sin(x) + \cos(x)\ln\abs{\cos(x)} + x\sin(x)$$

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

Problem Statement

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

$$\displaystyle{ y = c_1e^x + c_2xe^x + \frac{e^x}{6x^2} }$$

Problem Statement

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

Solution

### 607 solution video

video by Krista King Math

$$\displaystyle{ y = c_1e^x + c_2xe^x + \frac{e^x}{6x^2} }$$

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

Problem Statement

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

$$\displaystyle{ y = \frac{-2}{9}\cos(3x)\ln\abs{\sec(3x)+\tan(3x)} + A\cos(3x) + B\sin(3x) }$$

Problem Statement

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

Solution

### 602 solution video

video by Dr Chris Tisdell

$$\displaystyle{ y = \frac{-2}{9}\cos(3x)\ln\abs{\sec(3x)+\tan(3x)} + A\cos(3x) + B\sin(3x) }$$

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

Problem Statement

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

$$y(t) = (1/2)e^{-2t} - (1/2)e^{-t} + (1/2)\cos(t) + (3/2)\sin(t)$$

Problem Statement

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

Solution

### 605 solution video

video by Dr Chris Tisdell

$$y(t) = (1/2)e^{-2t} - (1/2)e^{-t} + (1/2)\cos(t) + (3/2)\sin(t)$$