Answer and Step-by-step explanation:
1. Data provided
[tex]\frac{d^2y}{dx^2} + 4y = x - x^2 + 20\\\\ \frac{d^2y}{dx^2} + 4y = - x^2 + x + 20[/tex]
Now as a non homogeneous part which is
[tex]- x^2 + x + 20[/tex] let us assume the computation is
[tex]y_p(x) = Ax^2 + Bx + C[/tex]
2. Data provided
[tex]\frac{ d^2y}{dx^2} + \frac{6dy}{dx} + 8y = e^{2x}[/tex]
As a non homogeneous part is [tex]e^2x[/tex] , let us assume the computation is
[tex]y_p(x) = Ae^{2x}[/tex]
3. Data provided
[tex]y'' + 4y' + 20y = -3sin2x[/tex]
As a non homogeneous part −3sin(2x), let us assume the computation is
[tex]y_p(x) = Acos(2x) + Bsin(2x)[/tex]
4. Data provided
[tex]y'' - 2y' - 15y = 3xcos(2x)[/tex]
As a non homogeneous part 3xcos(2x), let us assume the computation is
[tex]y_p(x) = (Ax+B)cos2x+(Cx+D)sin2x[/tex]
