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Mathematics LibreTexts

5.6E: Reduction of Order (Exercises)

  • Page ID
    18320
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    In Exercises [exer:5.6.1} –[exer:5.6.17} find the general solution, given that \(y_1\) satisfies the complementary equation. As a byproduct, find a fundamental set of solutions of the complementary equation.

    [exer:5.6.1] \((2x+1)y''-2y'-(2x+3)y=(2x+1)^2; \quad y_1=e^{-x}\)

    [exer:5.6.2] \(x^2y''+xy'-y={4\over x^2}; \quad y_1=x\)

    [exer:5.6.3] \(x^2y''-xy'+y=x; \quad y_1=x\)

    [exer:5.6.4] \(y''-3y'+2y={1\over1+e^{-x}}; \quad y_1=e^{2x}\)

    [exer:5.6.5] \(y''-2y'+y=7x^{3/2}e^x; \quad y_1=e^x\)

    [exer:5.6.6] \(4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=4x^{1/2}e^x(1+4x); \quad y_1=x^{1/2}e^x\)

    [exer:5.6.7] \(y''-2y'+2y=e^x\sec x; \quad y_1=e^x\cos x\)

    [exer:5.6.8] \(y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}; \quad y_1=e^{-x^2}\)

    [exer:5.6.9] \(x^2y''+xy'-4y=-6x-4; \quad y_1=x^2\)

    [exer:5.6.10] \(x^2y''+2x(x-1)y'+(x^2-2x+2)y=x^3e^{2x}; \quad y_1=xe^{-x}\)

    [exer:5.6.11] \(x^2y''-x(2x-1)y'+(x^2-x-1)y=x^2e^x; \quad y_1=xe^x\)

    [exer:5.6.12] \((1-2x)y''+2y'+(2x-3)y=(1-4x+4x^2)e^x; \quad y_1=e^x\)

    [exer:5.6.13] \(x^2y''-3xy'+4y=4x^4; \quad y_1=x^2\)

    [exer:5.6.14] \(2xy''+(4x+1)y'+(2x+1)y=3x^{1/2}e^{-x}; \quad y_1=e^{-x}\)

    [exer:5.6.15] \(xy''-(2x+1)y'+(x+1)y=-e^x; \quad y_1=e^x\)

    [exer:5.6.16] \(4x^2y''-4x(x+1)y'+(2x+3)y=4x^{5/2}e^{2x}; \quad y_1=x^{1/2}\)

    [exer:5.6.17] \(x^2y''-5xy'+8y=4x^2; \quad y_1=x^2\)

    [exer:5.6.18] \(xy''+(2-2x)y'+(x-2)y=0; \quad y_1=e^x\)

    [exer:5.6.19] \(x^2y''-4xy'+6y=0; \quad y_1=x^2\)

    [exer:5.6.20] \(x^2(\ln |x|)^2y''-(2x \ln |x|)y'+(2+\ln |x|)y=0; \quad y_1=\ln |x|\)

    [exer:5.6.21] \(4xy''+2y'+y=0; \quad y_1=\sin \sqrt{x}\)

    [exer:5.6.22] \(xy''-(2x+2)y'+(x+2)y=0; \quad y_1=e^x\)

    [exer:5.6.23] \(x^2y''-(2a-1)xy'+a^2y=0; \quad y_1=x^a\)

    [exer:5.6.24] \(x^2y''-2xy'+(x^2+2)y=0; \quad y_1=x \sin x\)

    [exer:5.6.25] \(xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}\)

    [exer:5.6.26] \(4x^2(\sin x)y''-4x(x\cos x+\sin x)y'+(2x\cos x+3\sin x)y=0; \quad y_1=x^{1/2}\)

    [exer:5.6.27] \(4x^2y''-4xy'+(3-16x^2)y=0; \quad y_1=x^{1/2}e^{2x}\)

    [exer:5.6.28] \((2x+1)xy''-2(2x^2-1)y'-4(x+1)y=0; \quad y_1=1/x\)

    [exer:5.6.29] \((x^2-2x)y''+(2-x^2)y'+(2x-2)y=0; \quad y_1=e^x\)

    [exer:5.6.30] \(xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}\)

    [exer:5.6.31] \(x^2y''-3xy'+4y=4x^4,\quad y(-1)=7,\quad y'(-1)=-8; \quad y_1=x^2\)

    [exer:5.6.32] \((3x-1)y''-(3x+2)y'-(6x-8)y=0, \quad y(0)=2,\; y'(0)=3; \quad y_1=e^{2x}\)

    [exer:5.6.33] \((x+1)^2y''-2(x+1)y'-(x^2+2x-1)y=(x+1)^3e^x, \quad y(0)=1,\quad y'(0)=~-1\);

    \(y_1=(x+1)e^x\)

    [exer:5.6.34] \(x^2y''+2xy'-2y=x^2, \quad y(1)={5\over4},\; y'(1)={3\over2}; \quad y_1=x\)

    [exer:5.6.35] \((x^2-4)y''+4xy'+2y=x+2, \quad y(0)=-{1\over3},\quad y'(0)=-1; \quad y_1={1\over x-2}\)

    [exer:5.6.36] Suppose \(p_1\) and \(p_2\) are continuous on \((a,b)\). Let \(y_1\) be a solution of

    \[y''+p_1(x)y'+p_2(x)y=0 \eqno{\rm (A)}\]

    that has no zeros on \((a,b)\), and let \(x_0\) be in \((a,b)\). Use reduction of order to show that \(y_1\) and

    \[y_2(x)=y_1(x)\int^x_{x_0}{1\over y^2_1(t)} \exp \left(-\int^t_{x_0}p_1(s)\, ds\right)\,dt\]

    form a fundamental set of solutions of (A) on \((a,b)\).

    [exer:5.6.37] The nonlinear first order equation

    \[y'+y^2+p(x)y+q(x)=0 \eqno{\rm (A)}\]

    is a Riccati equation. (See Exercise 2.4. [exer:2.4.55}.) Assume that \(p\) and \(q\) are continuous.

    Show that \(y\) is a solution of (A) if and only if \(y={z'/z}\), where

    \[z''+p(x)z'+q(x)z=0. \eqno{\rm (B)}\]

    Show that the general solution of (A) is

    \[y={c_1z'_1+c_2z'_2\over c_1z_1+c_2z_2}, \eqno{\rm (C)}\]

    where \(\{z_1,z_2\}\) is a fundamental set of solutions of (B) and \(c_1\) and \(c_2\) are arbitrary constants.

    Does the formula (C) imply that the first order equation (A) has a two–parameter family of solutions? Explain your answer.

    [exer:5.6.38] Use a method suggested by Exercise [exer:5.6.37} to find all solutions. of the equation.

    a

    \(y'+y^2+k^2=0\)

    b

    \(y'+y^2-3y+2=0\)

    c

    \(y'+y^2+5y-6=0\)

    d

    \(y'+y^2+8y+7=0\)

    e

    \(y'+y^2+14y+50=0\)

    f

    \(6y'+6y^2-y-1=0\)

    g

    \(36y'+36y^2-12y+1=0\)

    [exer:5.6.39] Use a method suggested by Exercise [exer:5.6.37} and reduction of order to find all solutions of the equation, given that \(y_1\) is a solution.

    \(x^2(y'+y^2)-x(x+2)y+x+2=0; \quad y_1=1/x\)

    \(y'+y^2+4xy+4x^2+2=0; \quad y_1=-2x\)

    \((2x+1)(y'+y^2)-2y-(2x+3)=0; \quad y_1=-1\)

    \((3x-1)(y'+y^2)-(3x+2)y-6x+8=0; \quad y_1=2\)

    \({x^2(y'+y^2)+xy+x^2- {1\over 4}=0; \quad y_1=-\tan x -{1\over 2x}}\)

    \({x^2(y'+y^2)-7xy+7=0; \quad y_1=1/x}\)

    [exer:5.6.40] The nonlinear first order equation

    \[y'+r(x)y^2+p(x)y+q(x)=0 \eqno{\rm (A)}\]

    is the generalized Riccati equation. (See Exercise 2.4. [exer:2.4.55}.) Assume that \(p\) and \(q\) are continuous and \(r\) is differentiable.

    Show that \(y\) is a solution of (A) if and only if \(y={z'/rz}\), where

    \[z''+\left[p(x)-{r'(x)\over r(x)}\right] z'+r(x)q(x)z=0. \eqno{\rm (B)}\]

    Show that the general solution of (A) is

    \[y={c_1z'_1+c_2z'_2\over r(c_1z_1+c_2z_2)},\]

    where \(\{z_1,z_2\}\) is a fundamental set of solutions of (B) and \(c_1\) and \(c_2\) are arbitrary constants.