Question

y"+9y=3sin3x 求微分方程的一个通解

Answer 1

问：y"+9y=3sin3x 求微分方程的一个通解

答：先解y''+ 9y = 0其特征方程为t² + 9 = 0t = ±3i则通解为y = C1cos3x + C2sin3x现在令y''+ 9y = sin3x有形如y = f(x)cos3x + g(x)sin3x的解,且令y' = f(x)[cos3x]' + g(x)[sin3x]',得到f'(x)cos3x + g'(x)sin3x = 0代入原式得到,- 3f'(x)sin3x - 9f(x)cos3x + 3g'(x)cos3x - 9g(x)sin3x + 9f(x)cos3x + 9g(x)sin3x = sin3x化简,- 3f'(x)sin3x + 3g'(x)cos3x = sin3x由f'(x)cos3x + g'(x)sin3x = 0联立解得f(x) = (sin6x)/36 - x/6 ,g(x) = -(cos6x)/36则其中一个特解为y = (sin6xcos3x)/36 - (xcos3x)/6 - (sin3xcos6x)/36 = (sin3x)/36 - (xcos3x)/6则原方程通解为y = C1cos3x + C2sin3x + (sin3x)/36 - (xcos3x)/6也可以写作y = C1cos3x + C2sin3x - (xcos3x)/6