Use the method of "undetermined coefficients" to solve the 2nd ODE #y''-2y'=12e^(2x)-8e^(-2x)#?

Question

Samantha Adkison · Accepted Answer

# y(x) = A + Be^(2x) + 6xe^(2x) - e^(-2x)#

We have: # y''-2y' = 12e^(2x) - 8e^(-2x) # This is a second order linear non-Homogeneous Differentiation Equation. The standard approach is to find a solution, #y_c# of the homogeneous equation by looking at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation. Complementary Role The equation homogeneous linked to [A] is # y''-2y'= 0 # Additionally, the related auxiliary equation is: # m^2-2m = 0 # # m(m-2) = 0 # Which has distinct real root #m=0# and #m=2# Consequently, the homogeneous equation's solution is: # y_c = Ae^(0x) + Be^(2x) # # \ \ \ = A + Be^(2x) # Specific Resolution With this particular equation [A], dur to the fact that #e^(2x)# is already part of the CF then a probable solution is of the form: # y = axe^(2x) + be^(-2x) # Where #a,b# are constant coefficients to be determined. Let us assume the above solution works, in which case be differentiating wrt #x# we have: # y' \ \ = 2axe^(2x)+ae^(2x)-2be^(-2x) # # \ \ \ \ \ = (2ax+a)e^(2x)-2be^(-2x) # # y'' = 2(2ax+a)e^(2x) + (2a)e^(2x)+4be^(-2x) # # \ \ \ \ \ = (4ax+4a)e^(2x) +4be^(-2x) # Substituting into the initial Differential Equation #[A]# we get: # {(4ax+4a)e^(2x) +4be^(-2x)} - 2{(2ax+a)e^(2x)-2be^(-2x)} = 12e^(2x) - 8e^(-2x) # # :. 4axe^(2x)+4ae^(2x) +4be^(-2x) - 4axe^(2x)-2ae^(2x)+4be^(-2x) = 12e^(2x) - 8e^(-2x) # # :. 4ae^(2x) +4be^(-2x) -2ae^(2x)+4be^(-2x) = 12e^(2x) - 8e^(-2x) # Equating coefficients of #e^(2x)# and #e^(-2x)# we get: # e^(2x) \ \ \ : 4a-2a=12 => a=6# # e^(-2x) : 4b+4b=-8 => b=-1 # Thus, we arrive at the specific solution: # y_p = 6xe^(2x) - e^(-2x) # Overall Resolution which ultimately results in the GS of [A} # y(x) = y_c + y_p # # \ \ \ \ \ \ \ = A + Be^(2x) + 6xe^(2x) - e^(-2x)#

Christopher Agresta · Answer

undefined To solve the second-order ordinary differential equation (ODE) $y'' - 2y' = 12e^{2x} - 8e^{-2x}$ using the method of undetermined coefficients, we first find the complementary function (CF) solution, then determine the particular integral (PI) solution.

1. Find the complementary function (CF) by solving the associated homogeneous equation:
$$y'' - 2y' = 0.$$
The characteristic equation is $r^2 - 2r = 0$, which has roots $r = 0$ and $r = 2$. Therefore, the CF is given by:
$$y_{CF} = c_1 + c_2 e^{2x},$$
where $c_1$ and $c_2$ are arbitrary constants.

2. Determine the particular integral (PI) by guessing a solution that matches the form of the nonhomogeneous term:
$$y_{PI} = Ae^{2x} + Be^{-2x},$$
where $A$ and $B$ are undetermined coefficients.

3. Substitute the guessed particular integral $y_{PI}$ into the original differential equation and solve for the coefficients $A$ and $B$.

$$y'' - 2y' = 12e^{2x} - 8e^{-2x}$$
$$2Ae^{2x} - 2Be^{-2x} - 4Ae^{2x} + 4Be^{-2x} = 12e^{2x} - 8e^{-2x}$$

4. Group the terms with the same exponential functions together and equate coefficients.

$$(-2A + 4B)e^{2x} + (2A + 4B)e^{-2x} = 12e^{2x} - 8e^{-2x}$$

Comparing coefficients:
$$(-2A + 4B) = 12$$ 
$$2A + 4B = -8$$

5. Solve the system of equations to find $A$ and $B$:
$$A = -4, \quad B = 2.$$

6. Write down the general solution:
$$y = y_{CF} + y_{PI} = c_1 + c_2 e^{2x} - 4e^{2x} + 2e^{-2x}.$$

This gives the general solution to the given second-order ordinary differential equation using the method of undetermined coefficients.