7.2 Nonhomogeneous linear equations (Page 5/9)

<< Chapter < Page

Calculus volume 3 Page 5 / 9

Chapter >> Page >

Using the method of variation of parameters

Find the general solution to the following differential equations.

$y ″ - 2 y^{'} + y = \frac{e^{t}}{t^{2}}$
$y ″ + y = 3 {sin}^{2} x$

The complementary equation is $y ″ - 2 y^{'} + y = 0$ with associated general solution $c_{1} e^{t} + c_{2} t e^{t} .$ Therefore, $y_{1} (t) = e^{t}$ and $y_{2} (t) = t e^{t} .$ Calculating the derivatives, we get $y_{1}^{'} (t) = e^{t}$ and $y_{2}^{'} (t) = e^{t} + t e^{t}$ (step 1). Then, we want to find functions $u^{'} (t)$ and $v^{'} (t)$ so that

$\begin{matrix} u^{'} e^{t} + v^{'} t e^{t} & = & 0 \\ u^{'} e^{t} + v^{'} (e^{t} + t e^{t}) & = & \frac{e^{t}}{t^{2}} . \end{matrix}$

Applying Cramer’s rule, we have

$u^{'} = \frac{| \begin{matrix} 0 & t e^{t} \\ \frac{e^{t}}{t^{2}} & e^{t} + t e^{t} \end{matrix} |}{| \begin{matrix} e^{t} & t e^{t} \\ e^{t} & e^{t} + t e^{t} \end{matrix} |} = \frac{0 - t e^{t} (\frac{e^{t}}{t^{2}})}{e^{t} (e^{t} + t e^{t}) - e^{t} t e^{t}} = \frac{- \frac{e^{2 t}}{t}}{e^{2 t}} = - \frac{1}{t}$

and

$v^{'} = \frac{| \begin{matrix} e^{t} & 0 \\ e^{t} & \frac{e^{t}}{t^{2}} \end{matrix} |}{| \begin{matrix} e^{t} & t e^{t} \\ e^{t} & e^{t} + t e^{t} \end{matrix} |} = \frac{e^{t} (\frac{e^{t}}{t^{2}})}{e^{2 t}} = \frac{1}{t^{2}} (step 2).$

Integrating, we get

$\begin{matrix} u = − \int \frac{1}{t} d t = − ln | t | \\ v = \int \frac{1}{t^{2}} d t = - \frac{1}{t} (step 3). \end{matrix}$

Then we have

$\begin{matrix} y_{p} & = − e^{t} ln | t | - \frac{1}{t} t e^{t} \\ = − e^{t} ln | t | - e^{t} (step 4). \end{matrix}$

The $e^{t}$ term is a solution to the complementary equation, so we don’t need to carry that term into our general solution explicitly. The general solution is

$y (t) = c_{1} e^{t} + c_{2} t e^{t} - e^{t} ln | t | (step 5).$
The complementary equation is $y ″ + y = 0$ with associated general solution $c_{1} cos x + c_{2} sin x .$ So, $y_{1} (x) = cos x$ and $y_{2} (x) = sin x$ (step 1). Then, we want to find functions $u^{'} (x)$ and $v^{'} (x)$ such that

$\begin{matrix} u^{'} cos x + v^{'} sin x & = & 0 \\ − u^{'} sin x + v^{'} cos x & = & 3 {sin}^{2} x . \end{matrix}$

Applying Cramer’s rule, we have

$u^{'} = \frac{| \begin{matrix} 0 & sin x \\ 3 {sin}^{2} x & cos x \end{matrix} |}{| \begin{matrix} cos x & sin x \\ − sin x & cos x \end{matrix} |} = \frac{0 - 3 {sin}^{3} x}{{cos}^{2} x + {sin}^{2} x} = −3 {sin}^{3} x$

and

$v^{'} = \frac{| \begin{matrix} cos x & 0 \\ − sin x & 3 {sin}^{2} x \end{matrix} |}{| \begin{matrix} cos x & sin x \\ − sin x & cos x \end{matrix} |} = \frac{3 {sin}^{2} x cos x}{1} = 3 {sin}^{2} x cos x (step 2).$

Integrating first to find u , we get

$u = \int −3 {sin}^{3} x d x = −3 [− \frac{1}{3} {sin}^{2} x cos x + \frac{2}{3} \int sin x d x] = {sin}^{2} x cos x + 2 cos x .$

Now, we integrate to find v . Using substitution (with $w = sin x$ ), we get

$v = \int 3 {sin}^{2} x cos x d x = \int 3 w^{2} d w = w^{3} = {sin}^{3} x .$

Then,

$\begin{matrix} y_{p} & = ({sin}^{2} x cos x + 2 cos x) cos x + ({sin}^{3} x) sin x \\ = {sin}^{2} x {cos}^{2} x + 2 {cos}^{2} x + {sin}^{4} x \\ = 2 {cos}^{2} x + {sin}^{2} x ({cos}^{2} x + {sin}^{2} x) (step 4). \\ = 2 {cos}^{2} x + {sin}^{2} x \\ = {cos}^{2} x + 1 \end{matrix}$

The general solution is

$y (x) = c_{1} cos x + c_{2} sin x + 1 + {cos}^{2} x (step 5).$

Got questions? Get instant answers now!

Find the general solution to the following differential equations.

$y ″ + y = sec x$
$x ″ - 2 x^{'} + x = \frac{e^{t}}{t}$

$y (x) = c_{1} cos x + c_{2} sin x + cos x ln | cos x | + x sin x$
$x (t) = c_{1} e^{t} + c_{2} t e^{t} + t e^{t} ln | t |$

Got questions? Get instant answers now!

Key concepts

To solve a nonhomogeneous linear second-order differential equation, first find the general solution to the complementary equation, then find a particular solution to the nonhomogeneous equation.
Let $y_{p} (x)$ be any particular solution to the nonhomogeneous linear differential equation

$a_{2} (x) y ″ + a_{1} (x) y^{'} + a_{0} (x) y = r (x),$

and let $c_{1} y_{1} (x) + c_{2} y_{2} (x)$ denote the general solution to the complementary equation. Then, the general solution to the nonhomogeneous equation is given by

$y (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x) + y_{p} (x).$
When $r (x)$ is a combination of polynomials, exponential functions, sines, and cosines, use the method of undetermined coefficients to find the particular solution. To use this method, assume a solution in the same form as $r (x),$ multiplying by x as necessary until the assumed solution is linearly independent of the general solution to the complementary equation. Then, substitute the assumed solution into the differential equation to find values for the coefficients.
When $r (x)$ is not a combination of polynomials, exponential functions, or sines and cosines, use the method of variation of parameters to find the particular solution. This method involves using Cramer’s rule or another suitable technique to find functions $u^{'} (x)$ and $v^{'} (x)$ satisfying

$\begin{matrix} u^{'} y_{1} + v^{'} y_{2} & = & 0 \\ u^{'} y_{1}^{'} + v^{'} y_{2}^{'} & = & r (x). \end{matrix}$

Then, $y_{p} (x) = u (x) y_{1} (x) + v (x) y_{2} (x)$ is a particular solution to the differential equation.

Key equations

Complementary equation
$a_{2} (x) y ″ + a_{1} (x) y^{'} + a_{0} (x) y = 0$
General solution to a nonhomogeneous linear differential equation
$y (x) = c_{1} y_{1} (x) + c_{2} y_{2} (x) + y_{p} (x)$

Solve the following equations using the method of undetermined coefficients.

$2 y ″ - 5 y^{'} - 12 y = 6$

Got questions? Get instant answers now!

$3 y ″ + y^{'} - 4 y = 8$

$y = c_{1} e^{−4 x / 3} + c_{2} e^{x} - 2$

Got questions? Get instant answers now!

$y ″ - 6 y^{'} + 5 y = e^{− x}$

Got questions? Get instant answers now!

$y ″ + 16 y = e^{−2 x}$

$y = c_{1} cos 4 x + c_{2} sin 4 x + \frac{1}{20} e^{−2 x}$

Got questions? Get instant answers now!

$y ″ - 4 y = x^{2} + 1$

Got questions? Get instant answers now!

$y ″ - 4 y^{'} + 4 y = 8 x^{2} + 4 x$

$y = c_{1} e^{2 x} + c_{2} x e^{2 x} + 2 x^{2} + 5 x$

Got questions? Get instant answers now!

$y ″ - 2 y^{'} - 3 y = sin 2 x$

Got questions? Get instant answers now!

$y ″ + 2 y^{'} + y = sin x + cos x$

$y = c_{1} e^{− x} + c_{2} x e^{− x} + \frac{1}{2} sin x - \frac{1}{2} cos x$

Got questions? Get instant answers now!

$y ″ + 9 y = e^{x} cos x$

Got questions? Get instant answers now!

$y ″ + y = 3 sin 2 x + x cos 2 x$

$y = c_{1} cos x + c_{2} sin x - \frac{1}{3} x cos 2 x - \frac{5}{9} sin 2 x$

Got questions? Get instant answers now!

$y ″ + 3 y^{'} - 28 y = 10 e^{4 x}$

Got questions? Get instant answers now!

$y ″ + 10 y^{'} + 25 y = x e^{−5 x} + 4$

$y = c_{1} e^{−5 x} + c_{2} x e^{−5 x} + \frac{1}{6} x^{3} e^{−5 x} + \frac{4}{25}$

Got questions? Get instant answers now!

In each of the following problems,

Write the form for the particular solution $y_{p} (x)$ for the method of undetermined coefficients.
[T] Use a computer algebra system to find a particular solution to the given equation.

$y ″ - y^{'} - y = x + e^{− x}$

Got questions? Get instant answers now!

$y ″ - 3 y = x^{2} - 4 x + 11$

a. $y_{p} (x) = A x^{2} + B x + C$
b. $y_{p} (x) = - \frac{1}{3} x^{2} + \frac{4}{3} x - \frac{35}{9}$

Got questions? Get instant answers now!

$y ″ - y^{'} - 4 y = e^{x} cos 3 x$

Got questions? Get instant answers now!

$2 y ″ - y^{'} + y = (x^{2} - 5 x) e^{− x}$

a. $y_{p} (x) = (A x^{2} + B x + C) e^{− x}$
b. $y_{p} (x) = (\frac{1}{4} x^{2} - \frac{5}{8} x - \frac{33}{32}) e^{− x}$

Got questions? Get instant answers now!

$4 y ″ + 5 y^{'} - 2 y = e^{2 x} + x sin x$

Got questions? Get instant answers now!

$y ″ - y^{'} - 2 y = x^{2} e^{x} sin x$

a. $y_{p} (x) = (A x^{2} + B x + C) e^{x} cos x$ $+ (D x^{2} + E x + F) e^{x} sin x$
b. $y_{p} (x) = (- \frac{1}{10} x^{2} - \frac{11}{25} x - \frac{27}{250}) e^{x} cos x$ $+ (- \frac{3}{10} x^{2} + \frac{2}{25} x + \frac{39}{250}) e^{x} sin x$

Got questions? Get instant answers now!

Solve the differential equation using either the method of undetermined coefficients or the variation of parameters.

$y ″ + 3 y^{'} - 4 y = 2 e^{x}$

Got questions? Get instant answers now!

$y ″ + 2 y^{'} = e^{3 x}$

$y = c_{1} + c_{2} e^{−2 x} + \frac{1}{15} e^{3 x}$

Got questions? Get instant answers now!

$y ″ + 6 y^{'} + 9 y = e^{− x}$

Got questions? Get instant answers now!

$y ″ + 2 y^{'} - 8 y = 6 e^{2 x}$

$y = c_{1} e^{2 x} + c_{2} e^{−4 x} + x e^{2 x}$

Got questions? Get instant answers now!

Solve the differential equation using the method of variation of parameters.

$4 y ″ + y = 2 sin x$

Got questions? Get instant answers now!

$y ″ - 9 y = 8 x$

$y = c_{1} e^{3 x} + c_{2} e^{−3 x} - \frac{8 x}{9}$

Got questions? Get instant answers now!

$y ″ + y = sec x, 0 < x < π / 2$

Got questions? Get instant answers now!

$y ″ + 4 y = 3 csc 2 x, 0 < x < π / 2$

$y = c_{1} cos 2 x + c_{2} sin 2 x - \frac{3}{2} x cos 2 x + \frac{3}{4} sin 2 x ln (sin 2 x)$

Got questions? Get instant answers now!

Find the unique solution satisfying the differential equation and the initial conditions given, where $y_{p} (x)$ is the particular solution.

$y ″ - 2 y^{'} + y = 12 e^{x},$ $y_{p} (x) = 6 x^{2} e^{x},$ $y (0) = 6, y' (0) = 0$

Got questions? Get instant answers now!

$y ″ - 7 y^{'} = 4 x e^{7 x},$ $y_{p} (x) = \frac{2}{7} x^{2} e^{7 x} - \frac{4}{49} x e^{7 x},$ $y (0) = −1, y' (0) = 0$

$y = - \frac{347}{343} + \frac{4}{343} e^{7 x} + \frac{2}{7} x^{2} e^{7 x} - \frac{4}{49} x e^{7 x}$

Got questions? Get instant answers now!

$y ″ + y = cos x - 4 sin x,$ $y_{p} (x) = 2 x cos x + \frac{1}{2} x sin x,$ $y (0) = 8, y' (0) = −4$

Got questions? Get instant answers now!

$y ″ - 5 y^{'} = e^{5 x} + 8 e^{−5 x},$ $y_{p} (x) = \frac{1}{5} x e^{5 x} + \frac{4}{25} e^{−5 x},$ $y (0) = −2, y' (0) = 0$

$y = - \frac{57}{25} + \frac{3}{25} e^{5 x} + \frac{1}{5} x e^{5 x} + \frac{4}{25} e^{−5 x}$

Got questions? Get instant answers now!

In each of the following problems, two linearly independent solutions— $y_{1}$ and $y_{2}$ —are given that satisfy the corresponding homogeneous equation. Use the method of variation of parameters to find a particular solution to the given nonhomogeneous equation. Assume x >0 in each exercise.

$x^{2} y ″ + 2 x y^{'} - 2 y = 3 x,$ $y_{1} (x) = x, y_{2} (x) = x^{−2}$

Got questions? Get instant answers now!

$x^{2} y ″ - 2 y = 10 x^{2} - 1,$ $y_{1} (x) = x^{2}, y_{2} (x) = x^{−1}$

$y_{p} = \frac{1}{2} + \frac{10}{3} x^{2} ln x$

Got questions? Get instant answers now!

<< Chapter < Page Page > Chapter >>

Practice Key Terms 4

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

100% Free Mobile Applications
Receive real-time job alerts and never miss the right job again

Source: OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2

Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Calculus volume 3' conversation and receive update notifications?

Ask

	Social Psychology MCQ By Saylor Foundation Start Quiz
	Anthropology Economic System By Richley Crapo Start Assignment
	3 Pharmacology Excl. Nervous System MCQ By Rohini Ajay Start Quiz
©flickr: Elliott	Spanish Verbs Subject Pronouns By Mariah Hauptman Start Quiz
	How to Analyze Stocks By Yasser Ibrahim Start Quiz
	Macroeconomics MCQ By Candice Butts Start Quiz
	Biology By OpenStax Read Online Course
	26 Toxicology Gustafson- Biotoxins and Venoms By Brooke Delaney Start Exam
	10 Psychology MCQ 2011 Final Exam By John Gabrieli Start Exam
	Social Work midterm By Katy Pratt Start Exam