5.2.1: Reduction of Order (Exercises)
- Page ID
- 103510
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In Exercises 1-30 find the general solution, given that \(y_1\) is a solution.
1. \((2x+1)y''-2y'-(2x+3)y=0; \quad y_1=e^{-x}\)
2. \(x^2y''+xy'-y=0; \quad y_1=x\)
3. \(x^2y''-xy'+y=0; \quad y_1=x\)
4. \(y''-3y'+2y=0; \quad y_1=e^{2x}\)
5. \(y''-2y'+y=0; \quad y_1=e^x\)
6. \(4x^2y''+(4x-8x^2)y'+(4x^2-4x-1)y=0; \quad y_1=x^{1/2}e^x\)
7. \(y''-2y'+2y=0; \quad y_1=e^x\cos x\)
8. \(y''+4xy'+(4x^2+2)y=0; \quad y_1=e^{-x^2}\)
9. \(x^2y''+xy'-4y=0; \quad y_1=x^2\)
10. \(x^2y''+2x(x-1)y'+(x^2-2x+2)y=0; \quad y_1=xe^{-x}\)
11. \(x^2y''-x(2x-1)y'+(x^2-x-1)y=0; \quad y_1=xe^x\)
12. \((1-2x)y''+2y'+(2x-3)y=0; \quad y_1=e^x\)
13. \(x^2y''-3xy'+4y=0; \quad y_1=x^2\)
14. \(2xy''+(4x+1)y'+(2x+1)y=0; \quad y_1=e^{-x}\)
15. \(xy''-(2x+1)y'+(x+1)y=0; \quad y_1=e^x\)
16. \(4x^2y''-4x(x+1)y'+(2x+3)y=0; \quad y_1=x^{1/2}\)
17. \(x^2y''-5xy'+8y=0; \quad y_1=x^2\)
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. \(xy''-(4x+1)y'+(4x+2)y=0; \quad y_1=e^{2x}\)
In Exercises 31-33 solve the initial value problem, given that \(y_{1}\) is a solution.
31. \(x^2y''-3xy'+4y=0,\quad y(-1)=7,\quad y'(-1)=-8; \quad y_1=x^2\)
32. \((x+1)^2y''-2(x+1)y'-(x^2+2x-1)y=0, \quad y(0)=1,\quad y'(0)=-1; \quad y_1=(x+1)e^x\)
33. \(x^2y''+2xy'-2y=0, \quad y(1)=0,\; y'(1)=3; \quad y_1=x\)