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Mathematics LibreTexts

5.7E: Variation of Parameters (Exercises)

  • Page ID
    18321
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    In Exercises [exer:5.7.1} –[exer:5.7.6} use variation of parameters to find a particular solution.

    [exer:5.7.1] \(y''+9y=\tan 3x\)

    [exer:5.7.2] \(y''+4y=\sin 2x\sec^2 2x\)

    [exer:5.7.3] \(y''-3y'+2y={4\over 1+e^{-x}}\)

    [exer:5.7.4] \(y''-2y'+2y=3e^x \sec x\) [exer:5.7.5] \(y''-2y'+y=14x^{3/2}e^x\)

    [exer:5.7.6] \(y''-y={4e^{-x}\over 1-e^{-2x}}\)

    [exer:5.7.7] \(x^2y''+xy'- y=2x^2+2; \quad y_1=x, \quad y_2={1\over x}\)

    [exer:5.7.8] \({xy''+(2-2x)y'+(x-2)y=e^{2x}; \quad y_1=e^x, \quad y_2={e^x\over x}}\)

    [exer:5.7.9] \(4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=4x^{1/2}e^x, \quad x > 0\); \(y_1=x^{1/2} e^x,\; y_2=x^{-1/2}e^x\)

    [exer:5.7.10] \(y''+4xy'+(4x^2+2)y=4e^{-x(x+2)};\quad y_1=e^{-x^2}, \quad y_2=xe^{-x^2}\)

    [exer:5.7.11] \(x^2y''-4xy'+6y=x^{5/2},\, x > 0;\quad y_1=x^2,\; y_2=x^3\)

    [exer:5.7.12] \(x^2y''-3xy'+3y=2x^4\sin x; \quad y_1=x,\; y_2=x^3\)

    [exer:5.7.13] \((2x+1)y''-2y'-(2x+3)y=(2x+1)^2e^{-x}; \quad y_1=e^{-x}, \quad y_2=xe^x\)

    [exer:5.7.14] \(4xy''+2y'+y=\sin\sqrt x; \quad y_1=\cos\sqrt x, \quad y_2=\sin\sqrt x\)

    [exer:5.7.15] \(xy''-(2x+2)y'+(x+2)y=6x^3e^x;\quad y_1=e^x,\quad y_2=x^3e^x\)

    [exer:5.7.16] \(x^2y''-(2a-1)xy'+a^2y=x^{a+1}; \quad y_1=x^a, \quad y_2=x^a \ln x\)

    [exer:5.7.17] \(x^2y''-2xy'+(x^2+2)y=x^3\cos x; \quad y_1=x\cos x, \quad y_2=x\sin x\)

    [exer:5.7.18] \(xy''-y'-4x^3y=8x^5;\quad y_1=e^{x^2},\; y_2=e^{-x^2}\)

    [exer:5.7.19] \((\sin x)y''+(2\sin x-\cos x)y'+(\sin x-\cos x)y=e^{-x}; \quad y_1=e^{-x},\quad y_2=e^{-x}\cos x\)

    [exer:5.7.20] \(4x^2y''-4xy'+(3-16x^2)y=8x^{5/2}; \quad y_1=\sqrt xe^{2x},\; y_2=\sqrt xe^{-2x}\)

    [exer:5.7.21] \(4x^2y''-4xy'+(4x^2+3)y=x^{7/2}; \quad y_1=\sqrt x\sin x,\; y_2=\sqrt x\cos x\)

    [exer:5.7.22] \(x^2y''-2xy'-(x^2-2)y=3x^4;\quad y_1=xe^x,\; y_2=xe^{-x}\)

    [exer:5.7.23] \(x^2y''-2x(x+1)y' +(x^2+2x+2)y=x^3e^x; \quad y_1=xe^x, \quad y_2=x^2e^x\)

    [exer:5.7.24] \(x^2y''-xy'-3y=x^{3/2}; \quad y_1=1/x, \quad y_2=x^3\)

    [exer:5.7.25] \(x^2y''-x(x+4)y'+2(x+3)y=x^4e^x; \quad y_1=x^2, \quad y_2=x^2e^x\)

    [exer:5.7.26] \(x^2y''-2x(x+2)y'+(x^2+4x+6)y=2xe^x; \quad y_1=x^2e^x, \quad y_2=x^3e^x\)

    [exer:5.7.27] \(x^2y''-4xy'+(x^2+6)y=x^4; \quad y_1=x^2\cos x, \quad y_2=x^2\sin x\)

    [exer:5.7.28] \((x-1)y''-xy'+y=2(x-1)^2e^x; \quad y_1=x, \quad y_2=e^x\)

    [exer:5.7.29] \(4x^2y''-4x(x+1)y'+(2x+3)y=x^{5/2}e^x; \quad y_1=\sqrt x, \quad y_2=\sqrt xe^x\)

    [exer:5.7.30] \((3x-1)y''-(3x+2)y'-(6x-8)y=(3x-1)^2e^{2x}, \quad y(0)=1,\; y'(0)=2\); \(y_1=e^{2x},\; y_2=xe^{-x}\)

    [exer:5.7.31] \((x-1)^2y''-2(x-1)y'+2y=(x-1)^2, \quad y(0)=3,\quad y'(0)=-6\);

    \(y_1=x-1\), \(y_2=x^2-1\)

    [exer:5.7.32] \((x-1)^2y''-(x^2-1)y'+(x+1)y=(x-1)^3e^x, \quad y(0)=4,\quad y'(0)=-6\);

    \(y_1=(x-1)e^x,\quad y_2=x-1\)

    [exer:5.7.33] \({(x^2-1)y''+4xy'+2y=2x, \quad y(0)=0,\; y'(0) =-2; \quad y_1={1\over x-1},\; y_2={1\over x+1}}\)

    [exer:5.7.34] \({x^2y''+2xy'-2y=-2x^2, \quad y(1)=1,\; y'(1)= -1; \quad y_1=x,\; y_2={1\over x^2}}\)

    [exer:5.7.35] \((x+1)(2x+3)y''+2(x+2)y'-2y=(2x+3)^2, \quad y(0)=0,\quad y'(0)=0\); \(y_1=x+2,\quad y_2={1\over x+1}\)

    [exer:5.7.36] Suppose

    \[y_p=\overline y+a_1y_1+a_2y_2\]

    is a particular solution of

    \[P_0(x)y''+P_1(x)y'+P_2(x)y=F(x), \eqno{\rm (A)}\]

    where \(y_1\) and \(y_2\) are solutions of the complementary equation

    \[P_0(x)y''+P_1(x)y'+P_2(x)y=0.\]

    Show that \(\overline y\) is also a solution of (A).

    [exer:5.7.37] Suppose \(p\), \(q\), and \(f\) are continuous on \((a,b)\) and let \(x_0\) be in \((a,b)\). Let \(y_1\) and \(y_2\) be the solutions of

    \[y''+p(x)y'+q(x)y=0\]

    such that

    \[y_1(x_0)=1, \quad y_1'(x_0)=0, \quad y_2(x_0)=0, \quad y_2'(x_0)=1.\]

    Use variation of parameters to show that the solution of the initial value problem

    \[y''+p(x)y'+q(x)y=f(x), \quad y(x_0)=k_0,\; y'(x_0)=k_1,\]

    is

    \[\begin{array}{rcl} y(x) &=& k_0y_1(x)+k_1y_2(x) \\ & & \qquad+\int^x_{x_0}\left(y_1(t)y_2(x)- y_1(x)y_2(t)\right) f(t)\exp\left(\int^t_{x_0}p(s)\,ds\right)\,dt. \end{array}\]

    Show also that

    \[\begin{array}{rcl} y'(x) &=& k_0y_1'(x)+k_1y_2'(x) \\[2\jot] & & \qquad+\int^x_{x_0}\left(y_1(t)y_2'(x)-y_1'(x)y_2(t) \right)f(t)\exp\left(\int^t_{x_0}p(s)\,ds\right)\,dt. \end{array}\]

    [exer:5.7.38] Suppose \(f\) is continuous on an open interval that contains \(x_0=0\). Use variation of parameters to find a formula for the solution of the initial value problem

    \[y''-y=f(x), \quad y(0)=k_0,\quad y'(0)=k_1.\]

    [exer:5.7.39] Suppose \(f\) is continuous on \((a,\infty)\), where \(a<0\), so \(x_0=0\) is in \((a,\infty)\).

    Use variation of parameters to find a formula for the solution of the initial value problem

    \[y''+y=f(x), \quad y(0)=k_0,\quad y'(0)=k_1.\]

    For the rest of this exercise assume that the improper integral \(\int_{0}^\infty f(t)\,dt\) is absolutely convergent.

    Show that if \(y\) is a solution of

    \[y''+y=f(x) \eqno{\rm (A)}\]

    on \((a,\infty)\), then

    \[\lim_{x \to \infty}\left(y(x)-A_0\cos x-A_1\sin x\right)=0 \eqno{\rm (B)}\]

    and

    \[\lim_{x\to\infty}\left(y'(x)+A_0\sin x-A_1\cos x\right)=0, \eqno{\rm (C)}\]

    where

    \[A_0=k_0-\int_0^\infty f(t)\sin t\,dt\mbox{\quad and \quad} A_1=k_1+\int_0^\infty f(t)\cos t\,dt.\]

    Show that if \(A_0\) and \(A_1\) are arbitrary constants, then there’s a unique solution of \(y''+y=f(x)\) on \((a,\infty)\) that satisfies (B) and (C).