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Mathematics LibreTexts

5.3.1: Nonhomgeneous Linear Equations (Exercises)

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    30727
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    Q5.3.1

    In Exercises 5.3.1-5.3.6 find a particular solution by the method used in Example 5.3.2. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

    1. \(y''+5y'-6y=22+18x-18x^2\)

    2. \(y''-4y'+5y=1+5x\)

    3. \(y''+8y'+7y=-8-x+24x^2+7x^3\)

    4. \(y''-4y'+4y=2+8x-4x^2\)

    5. \(y''+2y'+10y=4+26x+6x^2+10x^3, \quad y(0)=2, \quad y'(0)=9\)

    6. \(y''+6y'+10y=22+20x, \quad y(0)=2,\; y'(0)=-2\)

    Q5.3.2

    7. Show that the method used in Example 5.3.2 will not yield a particular solution of

    \[y''+y'=1+2x+x^2; \tag{A}\]

    that is, (A) does’nt have a particular solution of the form \(y_p=A+Bx+Cx^2\), where \(A\), \(B\), and \(C\) are constants.

    Q5.3.3

    In Exercises 5.3.8-5.3.13 find a particular solution by the method used in Example 5.3.3.

    8. \(x^{2}y'' +7xy'+8y=\frac{6}{x}\)

    9. \(x^{2}y''-7xy'+7y=13x^{1/2}\)

    10. \(x^{2}y''-xy'+y=2x^{3}\)

    11. \(x^{2}y''+5xy'+4y=\frac{1}{x^{3}}\)

    12. \(x^{2}y''+xy'+y=10x^{1/3}\)

    13. \(x^{2}y''-3xy'+13y=2x^{4}\)

    Q5.3.4

    14. Show that the method suggested for finding a particular solution in Exercises 5.3.8-5.3.13 will not yield a particular solution of

    \[x^2y''+3xy'-3y={1\over x^3}; \tag{A}\]

    that is, (A) doesn’t have a particular solution of the form \(y_p=A/x^3\).

    15. Prove: If \(a\), \(b\), \(c\), \(\alpha\), and \(M\) are constants and \(M\ne0\) then

    \[ax^2y''+bxy'+cy=M x^\alpha\]

    has a particular solution \(y_p=Ax^\alpha\) (\(A=\) constant) if and only if \(a\alpha(\alpha-1)+b\alpha+c\ne0\).

    Q5.3.5

    If \(a, b, c,\) and \(\alpha\) are constants, then \[\alpha (e^{\alpha x})'' +b(e^{\alpha x})'+ce^{\alpha x} = (a\alpha ^{2}+b\alpha + c)e^{\alpha x}.\] Use this in Exercises 5.3.16-5.3.21 to find a particular solution. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

    16. \(y''+5y'-6y=6e^{3x}\)

    17. \(y''-4y'+5y=e^{2x}\)

    18. \(y''+8y'+7y=10e^{-2x}, \quad y(0)=-2,\; y'(0)=10\)

    19. \(y''-4y'+4y=e^{x}, \quad y(0)=2,\quad y'(0)=0\)

    20. \(y''+2y'+10y=e^{x/2}\)

    21. \(y''+6y'+10y=e^{-3x}\)

    Q5.3.6

    22. Show that the method suggested for finding a particular solution in Exercises 5.3.16-5.3.21 will not yield a particular solution of

    \[y''-7y'+12y=5e^{4x}; \tag{A}\]

    that is, (A) doesn’t have a particular solution of the form \(y_p=Ae^{4x}\).

    23. Prove: If \(\alpha\) and \(M\) are constants and \(M\ne0\) then constant coefficient equation

    \[ay''+by'+cy=M e^{\alpha x}\]

    has a particular solution \(y_p=Ae^{\alpha x}\) (\(A=\) constant) if and only if \(e^{\alpha x}\) isn’t a solution of the complementary equation.

    Q5.3.7

    If \(ω\) is a constant, differentiating a linear combination of \(\cos ωx\) and \(\sin ωx\) with respect to \(x\) yields another linear combination of \(\cos ωx\) and \(\sin ωx\). In Exercises 5.3.24-5.3.29 use this to find a particular solution of the equation. Then find the general solution and, where indicated, solve the initial value problem and graph the solution.

    24. \(y''-8y'+16y=23\cos x-7\sin x\)

    25. \(y''+y'=-8\cos2x+6\sin2x\)

    26. \(y''-2y'+3y=-6\cos3x+6\sin3x\)

    27. \(y''+6y'+13y=18\cos x+6\sin x\)

    28. \(y''+7y'+12y=-2\cos2x+36\sin2x, \quad y(0)=-3,\quad y'(0)=3\)

    29. \(y''-6y'+9y=18\cos3x+18\sin3x, \quad y(0)=2,\quad y'(0)=2\)

    Q5.3.8

    30. Find the general solution of

    \[y''+\omega_0^2y =M\cos\omega x+N\sin\omega x,\]

    where \(M\) and \(N\) are constants and \(\omega\) and \(\omega_0\) are distinct positive numbers.

    31. Show that the method suggested for finding a particular solution in Exercises 5.3.24-5.3.29 will not yield a particular solution of

    \[y''+y=\cos x+\sin x; \tag{A}\]

    that is, (A) does not have a particular solution of the form \(y_p=A\cos x+B\sin x\).

    32. Prove: If \(M\), \(N\) are constants (not both zero) and \(\omega>0\), the constant coefficient equation

    \[ay''+by'+cy=M\cos\omega x+N\sin\omega x \tag{A}\]

    has a particular solution that’s a linear combination of \(\cos\omega x\) and \(\sin\omega x\) if and only if the left side of (A) is not of the form \(a(y''+\omega^2y)\), so that \(\cos\omega x\) and \(\sin\omega x\) are solutions of the complementary equation.

    Q5.3.9

    In Exercises 5.3.33-5.3.38 refer to the cited exercises and use the principal of superposition to find a particular solution. Then find the general solution.

    33. \(y''+5y'-6y=22+18x-18x^2+6e^{3x}\) (See Exercises 5.3.1 and 5.3.16.)

    34. \(y''-4y'+5y=1+5x+e^{2x}\) (See Exercises 5.3.2 and 5.3.17.)

    35. \(y''+8y'+7y=-8-x+24x^2+7x^3+10e^{-2x}\) (See Exercises 5.3.3 and 5.3.18.)

    36. \(y''-4y'+4y=2+8x-4x^2+e^{x}\) (See Exercises 5.3.4 and 5.3.19.)

    37. \(y''+2y'+10y=4+26x+6x^2+10x^3+e^{x/2}\) (See Exercises 5.3.5 and 5.3.20.)

    38. \(y''+6y'+10y=22+20x+e^{-3x}\) (See Exercises 5.3.6 and 5.3.21.)

    Q5.3.10

    39. Prove: If \(y_{p_1}\) is a particular solution of

    \[P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)\]

    on \((a,b)\) and \(y_{p_2}\) is a particular solution of

    \[P_0(x)y''+P_1(x)y'+P_2(x)y=F_2(x)\]

    on \((a,b)\), then \(y_p=y_{p_1}+y_{p_2}\) is a solution of

    \[P_0(x)y''+P_1(x)y'+P_2(x)y=F_1(x)+F_2(x)\]

    on \((a,b)\).

    40. Suppose \(p\), \(q\), and \(f\) are continuous on \((a,b)\). Let \(y_1\), \(y_2\), and \(y_p\) be twice differentiable on \((a,b)\), such that \(y=c_1y_1+c_2y_2+y_p\) is a solution of

    \[y''+p(x)y'+q(x)y=f\]

    on \((a,b)\) for every choice of the constants \(c_1,c_2\). Show that \(y_1\) and \(y_2\) are solutions of the complementary equation on \((a,b)\).