Question

i lution of the great the equation. PROBLEMS: Section 3.8 1/2 use the method of variation of parameters Brahim a parimar solution of the given nonhomogeneous equa- The found the general solution of the equation bytes 27+ y = 1 1 - = - y = 5e 14. xy + xy' - 4y = x(x + x) 1,(x) = x2 Y 2(x) = x-2 15. (1 - x)y" + xy' - y = 2(x - 1)2- y(x) = X Y 2(x) = et 1 16. y" + -y' + y = x-12 y, (x) = x 12 sin x y2(x) = x 12cos I 17. y - y = e- Y,(x) = et ta Y2(x) = e- C 6 y' + '= Why Didn't We Use This Method Before The perceptive reader might have asked why we didn't solution of the homogeneous equation to a particular when we solved the first-order equation y' + p(x) = flat in Chapter 2. We could have; however, the integrating method found both of these components at the same time. ertheless, for Problems 18-23, solve the given first-orderline equation both by the integrating factor method and by me the solution to the corresponding homogeneous equations particular solution of the nonhomogeneous equation. 18. y' = 1 y = c + x 19. y' + 2y = 1 y = ce 2*+; 7. y + y = sin r 1 - 2y + 2y = sin 9. y' + y = sec x tanx 10. y' + y = esc 11. y' - 3y' + 2y = 1 + 12 y + 2y' + y = e- ln x For Problems 13-17, find a particular solution of the nonhom- ogeneous equation, given that the functions y, (x) and y2(x) are linearly independent solutions of the corresponding how mogeneous equation. Note: The coefficient of y" must always bel, and hence a preliminary division may be required. 13. xºy - 2xy' + 2y = xsin x 81(x) = x Y2(x) = x2 THE EQUATION OF MOTION OF A MASS A SPRING 20. y' + 4y = et y = ce **

#14 please

Answer 1

2E Taking e yDy dz these alues Subathatiy gten de uni Hen as, cam be New Dy + e 22 4o D2 FD):0 22 + ce 2 2t -2t -2e C 2e A3 Wo are P 2 ** 2PCB) 2 2 e e 4E te -e + e (-42 + e -2E 42 2E 2 6 s

ez 16 2そ te 2乙 e 24 2 12 hence, general s ct m 2 2 16 12 4