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# 5.4E: The Method of Undetermined Coefficients I (Exercises)

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In Exercises [exer:5.4.1} –[exer:5.4.14} find a particular solution.

[exer:5.4.1] $$y''-3y'+2y=e^{3x}(1+x)$$

[exer:5.4.2] $$y''-6y'+5y=e^{-3x}(35-8x)$$

[exer:5.4.3] $$y''-2y'-3y=e^x(-8+3x)$$

[exer:5.4.4] $$y''+2y'+y=e^{2x}(-7-15x+9x^2)$$

[exer:5.4.5] $$y''+4y=e^{-x}(7-4x+5x^2)$$

[exer:5.4.6] $$y''-y'-2y=e^x(9+2x-4x^2)$$

[exer:5.4.7] $$y''-4y'-5y=-6xe^{-x}$$

[exer:5.4.8] $$y''-3y'+2y=e^x(3-4x)$$

[exer:5.4.9] $$y''+y'-12y=e^{3x}(-6+7x)$$

[exer:5.4.10] $$2y''-3y'-2y=e^{2x}(-6+10x)$$

[exer:5.4.11] $$y''+2y'+y=e^{-x}(2+3x)$$

[exer:5.4.12] $$y''-2y'+y=e^x(1-6x)$$

[exer:5.4.13] $$y''-4y'+4y=e^{2x}(1-3x+6x^2)$$

[exer:5.4.14] $$9y''+6y'+y=e^{-x/3}(2-4x+4x^2)$$

[exer:5.4.15] $$y''-3y'+2y=e^{3x}(1+x)$$

[exer:5.4.16] $$y''-6y'+8y=e^x(11-6x)$$

[exer:5.4.17] $$y''+6y'+9y=e^{2x}(3-5x)$$

[exer:5.4.18] $$y''+2y'-3y=-16xe^x$$

[exer:5.4.19] $$y''-2y'+y=e^x(2-12x)$$

[exer:5.4.20] $$y''-4y'-5y=9e^{2x}(1+x), \quad y(0)=0,\quad y'(0)=-10$$

[exer:5.4.21] $$y''+3y'-4y=e^{2x}(7+6x), \quad y(0)=2,\quad y'(0)=8$$

[exer:5.4.22] $$y''+4y'+3y=-e^{-x}(2+8x), \quad y(0)=1,\quad y'(0)=2$$

[exer:5.4.23] $$y''-3y'-10y=7e^{-2x}, \quad y(0)=1,\quad y'(0)=-17$$

[exer:5.4.24] $$y''+y'+y=xe^x+e^{-x}(1+2x)$$

[exer:5.4.25] $$y''-7y'+12y=-e^x(17-42x)-e^{3x}$$

[exer:5.4.26] $$y''-8y'+16y=6xe^{4x}+2+16x+16x^2$$

[exer:5.4.27] $$y''-3y'+2y=-e^{2x}(3+4x)-e^x$$

[exer:5.4.28] $$y''-2y'+2y=e^x(1+x)+e^{-x}(2-8x+5x^2)$$

[exer:5.4.29] $$y''+y=e^{-x}(2-4x+2x^2)+e^{3x}(8-12x-10x^2)$$

[exer:5.4.30]

Prove that $$y$$ is a solution of the constant coefficient equation

$ay''+by'+cy=e^{\alpha x}G(x) \eqno{\rm (A)}\nonumber$

if and only if $$y=ue^{\alpha x}$$, where $$u$$ satisfies

$au''+p'(\alpha)u'+p(\alpha)u=G(x) \eqno{\rm (B)}\nonumber$

and $$p(r)=ar^2+br+c$$ is the characteristic polynomial of the complementary equation

$ay''+by'+cy=0.\nonumber$

For the rest of this exercise, let $$G$$ be a polynomial. Give the requested proofs for the case where

$G(x)=g_0+g_1x+g_2x^2+g_3x^3.\nonumber$

Prove that if $$e^{\alpha x}$$ isn’t a solution of the complementary equation then (B) has a particular solution of the form $$u_p=A(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, as in Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.4} . Conclude that (A) has a particular solution of the form $$y_p=e^{\alpha x}A(x)$$.

Show that if $$e^{\alpha x}$$ is a solution of the complementary equation and $$xe^{\alpha x}$$ isn’t , then (B) has a particular solution of the form $$u_p=xA(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, as in Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.5} . Conclude that (A) has a particular solution of the form $$y_p=xe^{\alpha x}A(x)$$.

Show that if $$e^{\alpha x}$$ and $$xe^{\alpha x}$$ are both solutions of the complementary equation then (B) has a particular solution of the form $$u_p=x^2A(x)$$, where $$A$$ is a polynomial of the same degree as $$G$$, and $$x^2A(x)$$ can be obtained by integrating $$G/a$$ twice, taking the constants of integration to be zero, as in Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.6} . Conclude that (A) has a particular solution of the form $$y_p=x^2e^{\alpha x}A(x)$$.

[exer:5.4.31] Compare with Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.1} :

$y''-7y'+12y=4e^{2x};\quad y_p=Ae^{2x}\nonumber$

[exer:5.4.32] Compare with Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.2} :

$y''-7y'+12y=5e^{4x};\quad y_p=Axe^{4x}\nonumber$

[exer:5.4.33] Compare with Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.3} .

$y''-8y'+16y=2e^{4x};\quad y_p=Ax^2e^{4x}\nonumber$

[exer:5.4.34] Compare with Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.4} :

$y''-3y'+2y=e^{3x}(-1+2x+x^2),\quad y_p=e^{3x}(A+Bx+Cx^2)\nonumber$

[exer:5.4.35] Compare with Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.5} :

$y''-4y'+3y=e^{3x}(6+8x+12x^2),\quad y_p=e^{3x}(Ax+Bx^2+Cx^3)\nonumber$

[exer:5.4.36] Compare with Example

Example $$\PageIndex{1}$$:

Add text here. For the automatic number to work, you need to add the “AutoNum” template (preferably at5.4.6} :

$4y''+4y'+y=e^{-x/2}(-8+48x+144x^2),\quad y_p=e^{-x/2}(Ax^2+Bx^3+Cx^4)\nonumber$

[exer:5.4.37] Write $$y=ue^{\alpha x}$$ to find the general solution.

1. $$y''+2y'+y={e^{-x}\over\sqrt x}$$
2. $$y''+6y'+9y=e^{-3x}\ln x$$
3. $$y''-4y'+4y={e^{2x}\over1+x}$$
4. $$4y''+4y'+y={4e^{-x/2}\left({1\over x}+x\right)}$$

[exer:5.4.38] Suppose $$\alpha\ne0$$ and $$k$$ is a positive integer. In most calculus books integrals like $$\int x^k e^{\alpha x}\,dx$$ are evaluated by integrating by parts $$k$$ times. This exercise presents another method. Let

$y=\int e^{\alpha x}P(x)\,dx\nonumber$

with

$P(x)=p_0+p_1x+\cdots+p_kx^k\nonumber$

(where $$p_k \neq 0$$).

Show that $$y=e^{\alpha x}u$$, where

$u'+\alpha u=P(x). \eqno{\rm (A)}\nonumber$

Show that (A) has a particular solution of the form

$u_p=A_0+A_1x+\cdots+A_kx^k,\nonumber$

where $$A_k$$, $$A_{k-1}$$, …, $$A_0$$ can be computed successively by equating coefficients of $$x^k,x^{k-1}, \dots,1$$ on both sides of the equation

$u_p'+\alpha u_p=P(x).\nonumber$

Conclude that

$\int e^{\alpha x}P(x)\,dx=\left(A_0+A_1x+\cdots+A_kx^k\right)e^{\alpha x} +c,\nonumber$

where $$c$$ is a constant of integration.

[exer:5.4.39] Use the method of Exercise [exer:5.4.38} to evaluate the integral.

## a

$$\int e^x(4+x)\,dx$$

## b

$$\int e^{-x}(-1+x^2)\,dx$$

[10pt]

## c

$$\int x^3e^{-2x}\,dx$$

## d

$$\int e^x(1+x)^2\,dx$$

[10pt]

## e

$$\int e^{3x}(-14+30x+27x^2)\,dx$$

## f

$$\int e^{-x}(1+6x^2-14x^3+3x^4)\,dx$$

[exer:5.4.40] Use the method suggested in Exercise [exer:5.4.38} to evaluate $$\int x^ke^{\alpha x}\,dx$$, where $$k$$ is an arbitrary positive integer and $$\alpha\ne0$$.