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7.1.1: Introduction to Linear Higher Order Equations (Exercises)

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In Exercises 1-6 verify that the given function is the solution of the initial value problem.

1. Verify that y=6x8x23x3+1x satisfies the IVP:

x3y3x2y+6xy6y=24x;y(1)=0,y(1)=0,y(1)=0

2. Verify that y=x+12x satisfies the IVP:

y1xyy+1xy=x24x4,y(1)=32,y(1)=12,y(1)=1

3. Verify that y=x22+2e(x1)e(x1)+4x satisfies the IVP:

xyyxy+y=x2,y(1)=2,y(1)=5,y(1)=1

4. Verify that y=2x2lnxx1/2+2x1/2+4x2 satisfies the IVP:

4x3y+4x2y5xy+2y=30x2,y(1)=5,y(1)=172,y(1)=634

5. Verify that y=x4lnx+x2x2+3x34x4 satisfies the IVP

x4y(4)4x3y+12x2y24xy+24y=6x4,y(1)=2,y(1)=9,y(1)=27,y(1)=52

6. Verify that y=912x+6x28x3 satisfies the IVP:

xy(4)y4xy+4y=96x2,y(1)=5,y(1)=24,y(1)=36;y(1)=48

7. Using the solution from Example 7.1.1, solve the initial value problem

x3yx2y2xy+6y=0,y(1)=4,y(1)=14,y(1)=20.

8. Using the solution from Example 7.1.2, solve the initial value problem

y(4)+y7yy+6y=0,y(0)=5,y(0)=6,y(0)=10,y(0)36.

9.

  1. Verify that the function y=c1x3+c2x2+c3x satisfies x3yx2y2xy+6y=0 if c1, c2, and c3 are constants.
  2. Use (a) to find solutions y1, y2, and y3 of (A) such that y1(1)=1,y1(1)=0,y1(1)=0y2(1)=0,y2(1)=1,y2(1)=0y3(1)=0,y3(1)=0,y3(1)=1.

In Exercises 10-16 verify that the given functions are solutions of the given equation, and show that they form a fundamental set of solutions of the equation on any interval on which the equation has continuous coefficients.

10. y+yyy=0;{ex,ex,xex}

11. y3y+7y5y=0;{ex,excos2x,exsin2x}

12. xyyxy+y=0;{ex,ex,x}

13. x2y+2xy(x2+2)y=0;{ex/x,ex/x,1}

14. (x22x+2)yx2y+2xy2y=0;{x,x2,ex}

15. (2x1)y(4)4xy+(52x)y+4xy4y=0;{x,ex,ex,e2x}

16. xy(4)y4xy+4y=0;{1,x2,e2x,e2x}

In Exercise 17-27 compute the Wronskian of the given set of functions and what your result says about the linear independence of the set.

17. {1,ex,ex}

18. {ex,exsinx,excosx}

19. {2,x+1,x2+2}

20. {x,xlnx,1/x}

21. {1,x,x22!,x33!,,xnn!}

22. {ex,ex,x}

23. {ex/x,ex/x,1}

24. {x,x2,ex}

25. {x,x3,1/x,1/x2}

26. {ex,ex,x,e2x}

27. {e2x,e2x,1,x2}

28. Suppose y(n)+p1(x)y(n1)++pn(x)y=f(x) has continuous coefficients on (a,b) and x0 is in (a,b). Use Theorem 7.1.1 to prove that y0 is the only solution of the initial value problem Ly=0,y(x0)=0,y(x0)=0,,y(n1)(x0)=0, on (a,b).

29. Prove: If y=c1y1+c2y2++ckyk+yp is a solution of a linear equation y(n)+p1(x)y(n1)++pn(x)y=f(x) for every choice of the constants c1, c2,…, ck, then yi is a solution to y(n)+p1(x)y(n1)++pn(x)y=0 for 1ik.


This page titled 7.1.1: Introduction to Linear Higher Order Equations (Exercises) is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform.

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