What is the general solution of the differential equation? : # (d^2y)/(dx^2)- y = 1/(1+e^x) #

Question

Emily Ammerman · Accepted Answer

# y(x) = Ae^x+Be^(-x) -1/2xe^x - 1/2 - 1/2e^xln(1+e^x) -1/2 e^-x ln(1+e^x)#

We have: # (d^2y)/(dx^2)- y = 1/(1+e^x) # ..... [A] 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 Function The homogeneous equation associated with [A] is # y'' -y = 0 # And it's associated Auxiliary equation is: # m^2 -1 = 0 # Which has distinct real solutions #m=+-1# Thus the solution of the homogeneous equation is: # y_c = Ae^x+Be^(-x) # Particular Solution With this particular equation [A], the interesting part is find the solution of the particular function. We would typically use practice & experience to "guess" the form of the solution but that approach is likely to fail here. Instead we must use the Wronskian (or variation of parameters). It does, however, involve a lot more work: Once we have two linearly independent solutions say #y_1(x)# and #y_2(x)# then the particular solution of the general DE; # ay'' +by' + cy = p(x) # is given by: # y_p = v_1y_1 + v_2y_2 \ \ #, which are all functions of #x# Where: # v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx # # v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx # And, #W[y_1,y_2]# is the wronskian; defined by the following determinant: # W[y_1,y_2] = | ( y_1,y_2), (y_1',y_2') | # So for our equation [A]: # p(x) = 1/(1+e^x) # # y_1 \ \ \ = e^x \ \ \ => y_1' = e^x # # y_2 \ \ \ = e^(-x) => y_2' = -e^-x # So the wronskian for this equation is: # W[y_1,y_2] = | ( e^x,,e^-x), (e^x,,-e^-x) | # # " " = (e^x)(-e^-x) - (e^x)(e^-x) # # " " = -2 # So we form the two particular solution function: # v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx # # \ \ \ = -int \ (1/(1+e^x) (e^-x))/(-2) \ dx # # \ \ \ = 1/2 \ int \ e^-x/(1+e^x) \ dx # We can evaluate this integral using a substitution: Let #u=e^x => (du)/dx = e^x # and #e^(-x)=1/u# then # v_1 = 1/2 \ int \ (1/u)/(1+u) \ (1/u) \ du# # \ \ \ = 1/2 \ int \ 1/(u^2(1+u)) \ du# No we decompose the integrand into partial fractions: # 1/(u^2(1+u)) -= a/u + b/u^2+c/(1+u) # # " " = (au(1+u) + b(1+u) + c u^2) / (u^2(1+u)) # # => 1 -= au(1+u) + b(1+u) + c u^2 # Where #a,b,c# are constants to be determine. Using substitution we have: Put # u=0 => 1 = b # Put # u=-11 => 1 = c # Coeff#(u^2): 0 = a+c => a=-1 # Thus: # v_1 = 1/2 \ int \ -1/u + 1/u^2+1/(1+u) \ du# # \ \ \ = 1/2 \ { -ln|u| - 1/u+ln|1+u| }# # \ \ \ = 1/2 \ { -ln|e^x| - 1/e^x+ln|1+e^x| }# # \ \ \ = 1/2 \ { -x - e^-x+ln(1+e^x) }# # \ \ \ = -1/2x - 1/2e^-x + 1/2ln(1+e^x) # And; # v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx # # \ \ \ = int \ (1/(1+e^x) e^x)/(-2) \ dx # # \ \ \ = -1/2 \ int \ (e^x)/(1+e^x) \ dx # We can similarly evaluate this integral using a substitution: Let #u=e^x => (du)/dx = e^x # then # v_2 = -1/2 \ int \ (u)/(1+u) \ (1/u) \ du # # \ \ \ = -1/2 \ int \ (u)/(1+u) \ (1/u) \ du # # \ \ \ = -1/2 \ int \ (1)/(1+u) \ du # # \ \ \ = -1/2 \ ln|1+u| # # \ \ \ = -1/2 \ ln|1+e^x| # # \ \ \ = -1/2 \ ln(1+e^x) # And so we form the Particular solution: # y_p = v_1y_1 + v_2y_2 # # \ \ \ = (-1/2x - 1/2e^-x + 1/2ln(1+e^x))(e^x) + (-1/2 \ ln(1+e^x))(e^-x)# # \ \ \ = -1/2xe^x - 1/2 - 1/2e^xln(1+e^x) -1/2 e^-x ln(1+e^x)# Which then leads to the GS of [A} # y(x) = y_c + y_p # # \ \ \ \ \ \ \ = Ae^x+Be^(-x) -1/2xe^x - 1/2 - 1/2e^xln(1+e^x) -1/2 e^-x ln(1+e^x)#