5.6E: Reduction of Order (Exercises)

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Q5.6.1

In Exercises 5.6.1-5.6.17 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.

1. \((2x+1)y''-2y'-(2x+3)y=(2x+1)^2; \quad y_1=e^{-x}\)

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

3. \(x^2y''-xy'+y=x; \quad y_1=x\)

4. \(y''-3y'+2y={1\over1+e^{-x}}; \quad y_1=e^{2x}\)

5. \(y''-2y'+y=7x^{3/2}e^x; \quad y_1=e^x\)

6. \(4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=4x^{1/2}e^x(1+4x); \quad y_1=x^{1/2}e^x\)

7. \(y''-2y'+2y=e^x\sec x; \quad y_1=e^x\cos x\)

8. \(y''+4xy'+(4x^2+2)y=8e^{-x(x+2)}; \quad y_1=e^{-x^2}\)

9. \(x^2y''+xy'-4y=-6x-4; \quad y_1=x^2\)

10. \(x^2y''+2x(x-1)y'+(x^2-2x+2)y=x^3e^{2x}; \quad y_1=xe^{-x}\)

11. \(x^2y''-x(2x-1)y'+(x^2-x-1)y=x^2e^x; \quad y_1=xe^x\)

12. \((1-2x)y''+2y'+(2x-3)y=(1-4x+4x^2)e^x; \quad y_1=e^x\)

13. \(x^2y''-3xy'+4y=4x^4; \quad y_1=x^2\)

14. \(2xy''+(4x+1)y'+(2x+1)y=3x^{1/2}e^{-x}; \quad y_1=e^{-x}\)

15. \(xy''-(2x+1)y'+(x+1)y=-e^x; \quad y_1=e^x\)

16. \(4x^2y''-4x(x+1)y'+(2x+3)y=4x^{5/2}e^{2x}; \quad y_1=x^{1/2}\)

17. \(x^2y''-5xy'+8y=4x^2; \quad y_1=x^2\)

Q5.6.2

In Exercises 5.6.18-5.6.30 find a fundamental set of solutions, given that \(y_{1}\) is a solution.

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

19. \(x^2y''-4xy'+6y=0; \quad y_1=x^2\)

20. \(x^2(\ln |x|)^2y''-(2x \ln |x|)y'+(2+\ln |x|)y=0; \quad y_1=\ln |x|\)

21. \(4xy''+2y'+y=0; \quad y_1=\sin \sqrt{x}\)

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

23. \(x^2y''-(2a-1)xy'+a^2y=0; \quad y_1=x^a\)

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

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

26. \(4x^2(\sin x)y''-4x(x\cos x+\sin x)y'+(2x\cos x+3\sin x)y=0; \quad y_1=x^{1/2}\)

27. \(4x^2y''-4xy'+(3-16x^2)y=0; \quad y_1=x^{1/2}e^{2x}\)

28. \((2x+1)xy''-2(2x^2-1)y'-4(x+1)y=0; \quad y_1=1/x\)

29. \((x^2-2x)y''+(2-x^2)y'+(2x-2)y=0; \quad y_1=e^x\)

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

Q5.6.3

In Exercises 5.6.31-5.6.33 solve the initial value problem, given that \(y_{1}\) satisfies the complementary equation.

31. \(x^2y''-3xy'+4y=4x^4,\quad y(-1)=7,\quad y'(-1)=-8; \quad y_1=x^2\)

32. \((3x-1)y''-(3x+2)y'-(6x-8)y=0, \quad y(0)=2,\; y'(0)=3; \quad y_1=e^{2x}\)

33. \((x+1)^2y''-2(x+1)y'-(x^2+2x-1)y=(x+1)^3e^x, \quad y(0)=1,\quad y'(0)=~-1; \quad y_1=(x+1)e^x\)

Q5.6.4

In Exercises 5.6.34 and 5.6.35 solve the initial value problem and graph the solution, given that \(y_{1}\) satisfies the complementary equation.

34. \(x^2y''+2xy'-2y=x^2, \quad y(1)={5\over4},\; y'(1)={3\over2}; \quad y_1=x\)

35. \((x^2-4)y''+4xy'+2y=x+2, \quad y(0)=-{1\over3},\quad y'(0)=-1; \quad y_1={1\over x-2}\)

Q5.6.5

36. Suppose \(p_1\) and \(p_2\) are continuous on \((a,b)\). Let \(y_1\) be a solution of

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

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 \((a,b)\).

37. The nonlinear first order equation

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

is a Riccati equation. (See Exercise 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 \[z''+p(x)z'+q(x)z=0. \tag{B}\]
Show that the general solution of (A) is \[y={c_1z'_1+c_2z'_2\over c_1z_1+c_2z_2}, \tag{C}\] 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.

38. Use a method suggested by Exercise 5.6.37 to find all solutions. of the equation.

\(y'+y^2+k^2=0\)
\(y'+y^2-3y+2=0\)
\(y'+y^2+5y-6=0\)
\(y'+y^2+8y+7=0\)
\(y'+y^2+14y+50=0\)
\(6y'+6y^2-y-1=0\)
\(36y'+36y^2-12y+1=0\)

39. Use a method suggested by Exercise 5.6.37 and reduction of order to find all solutions of the equation, given that \(y_1\) is a solution.

\(x^2(y'+y^2)-x(x+2)y+x+2=0; \quad y_1=1/x\)
\(y'+y^2+4xy+4x^2+2=0; \quad y_1=-2x\)
\((2x+1)(y'+y^2)-2y-(2x+3)=0; \quad y_1=-1\)
\((3x-1)(y'+y^2)-(3x+2)y-6x+8=0; \quad y_1=2\)
\({x^2(y'+y^2)+xy+x^2- {1\over 4}=0; \quad y_1=-\tan x -{1\over 2x}}\)
\({x^2(y'+y^2)-7xy+7=0; \quad y_1=1/x}\)

40. The nonlinear first order equation

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

is the generalized Riccati equation. (See Exercise 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 \[z''+\left[p(x)-{r'(x)\over r(x)}\right] z'+r(x)q(x)z=0. \tag{B}\]
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.