Final answer:
To solve the given differential equation by undetermined coefficients, we first need to find the complementary solution to the homogeneous equation. Next, we need to find the particular solution to the non-homogeneous equation by assuming a form for the particular solution and solving for the coefficients. The general solution is the sum of the complementary solution and the particular solution.
Step-by-step explanation:
To solve the given differential equation by undetermined coefficients, we first need to find the complementary solution to the homogeneous equation. The homogeneous equation is y′′′+y′′=0. The characteristic equation for this equation is r^3+r^2=0. Solving this equation, we have r^2(r+1)=0, which gives us the solutions r=0 and r=-1. Therefore, the homogeneous solution is y_h=C_1+C_2e^{0x}+C_3e^{-1x}=C_1+C_2+C_3e^{-x}, where C_1, C_2, and C_3 are constants.
Next, we need to find the particular solution to the non-homogeneous equation y′′′+y′′=6x^2. Since the right-hand side is a polynomial of degree 2, we assume the particular solution has the form y_p=Ax^2+Bx+C, where A, B, and C are constants to be determined.
By plugging this particular solution into the non-homogeneous equation, we can solve for the coefficients A, B, and C. Differentiating y_p three times, we have y_p''''=2A. Substituting this into the non-homogeneous equation, we get 2A+2A=6x^2, which implies A=3x^2/4.
The particular solution is y_p=(3x^2/4)x^2+Bx+C. The general solution is y=y_h+y_p=C_1+C_2+C_3e^{-x}+3x^2/4x^2+Bx+C, where C_1, C_2, C_3, B, and C are constants.