If a particular integral of the differential equation #(D^2+2D-1)y=e^(ax)# is #(-4/7)e^(ax)# then the value of a is ?

Question

William Abdalla · Accepted Answer

# a=-3/2, -1/2# #
We cannot eliminate a solution, thus we are left with two possibilities

# y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-3/2x) #

or

# y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-1/2x) #
We have: # (D^2+2D-1)y = e^(ax) # with a PI, #-4/7e^(ax) # Alternatively, in standard form: # y'' + 2y' -y = e^(ax) # This is a second order linear non-Homogeneous Differentiation Equation with constant coefficients. 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'-y = 0# Additionally, the related auxiliary equation is: # m^2+2m-1 = 0 => (m+1)^2-1-1 = 0# Which has two real and distinct solution #m=-1+-sqrt(2) # Consequently, the homogeneous equation's solution is: # y_c = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) # Specific Resolution We would search for a solution of the following form in order to identify a specific solution of the non-homogeneous equation: # y = Ke^(ax) # Where the constants #K# is to be determined by direct substitution and comparison: Differentiating wrt #x# we get: # \ y' = aKe^(ax) # # y'' = a^2Ke^(ax) # When we replace the DE [A] with: # (a^2Ke^(ax)) + 2(aKe^(ax)) - (Ke^(ax)) = e^(ax) # # :. K(a^2 + 2a - 1) = 1 # # :. K = 1/(a^2 + 2a - 1) # We are also given that the PS is #-4/7e^(ax)# # -4/7e^(ax) = Ke^(ax) => K=-4/7# Allowing us to find #a# using: # 1/(a^2 + 2a - 1) = -4/7 # # :. 4(a^2 + 2a - 1) = -7 # # :. 4a^2 + 8a +3=0 # # :. (2a+1)(2a+3) = 0 # # :. a=-3/2, -1/2# # We cannot eliminate a solution, thus we are left with two possibilities, Allowing us to write the General Solution # y = y_c + y_p # as: # y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-3/2x) # or # y = Ae^((-1-sqrt(2))x) + Be^((-1+sqrt(2))x) -4/7e^(-1/2x) #

Christopher Agresta · Answer

undefined To find the value of $ a $ in the given differential equation $ (D^2+2D-1)y=e^{ax} $ when the particular integral is $ (-\frac{4}{7})e^{ax} $, we can use the method of undetermined coefficients. By substituting $ y_p = Ae^{ax} $ into the differential equation, we can determine the value of $ a $. Comparing coefficients, we find that $ a = 1 $.