7.2 Nonhomogeneous linear equations (Page 5/9)

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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

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

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Find the general solution to the following differential equations.

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

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

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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.

$2 y ″ - 5 y^{'} - 12 y = 6$

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$3 y ″ + y^{'} - 4 y = 8$

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

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$y ″ - 6 y^{'} + 5 y = e^{− x}$

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$y ″ + 16 y = e^{−2 x}$

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

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$y ″ - 4 y = x^{2} + 1$

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$y ″ - 4 y^{'} + 4 y = 8 x^{2} + 4 x$

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

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$y ″ - 2 y^{'} - 3 y = sin 2 x$

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$y ″ + 2 y^{'} + y = sin x + cos x$

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

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$y ″ + 9 y = e^{x} cos x$

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$y ″ + y = 3 sin 2 x + x cos 2 x$

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

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$y ″ + 3 y^{'} - 28 y = 10 e^{4 x}$

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

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

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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.

$y ″ - y^{'} - y = x + e^{− x}$

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$y ″ - 3 y = x^{2} - 4 x + 11$

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

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$y ″ - y^{'} - 4 y = e^{x} cos 3 x$

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

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

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

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

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. $y_{p} (x) = (- \frac{1}{10} x^{2} - \frac{11}{25} x - \frac{27}{250}) e^{x} cos x$ $+ (- \frac{3}{10} x^{2} + \frac{2}{25} x + \frac{39}{250}) e^{x} sin x$

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

$y ″ + 3 y^{'} - 4 y = 2 e^{x}$

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$y ″ + 2 y^{'} = e^{3 x}$

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

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$y ″ + 6 y^{'} + 9 y = e^{− x}$

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$y ″ + 2 y^{'} - 8 y = 6 e^{2 x}$

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

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

$4 y ″ + y = 2 sin x$

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$y ″ - 9 y = 8 x$

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

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$y ″ + y = sec x, 0 < x < π / 2$

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

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

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

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

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

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

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

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

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

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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.

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

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

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

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

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