What is the general solution of the differential equation? : # Acosx + Bsinx -sinxln|cscx+cotx| #

Question

Emilia Aguilar · Accepted Answer

# Acosx + Bsinx -sinxln|cscx+cotx| #

We have: # y'' + y = cot(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'' + 9y = 0 # And it's associated Auxiliary equation is: # m^2 + 1 = 0 # Which has pure imaginary solutions #m=+-i# Thus the solution of the homogeneous equation is: # y_c = e^0(Acos(1x) + Bsin(1x)) # # \ \ \ = Acosx + Bsinx # 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. 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) = cotx # # y_1 \ \ \ = cosx => y'_1 = -sinx # # y_2 \ \ \ = sinx => y'_2 = cosx # So the wronskian for this equation is: # W[y_1,y_2] = | ( cosx,,sinx), (-sinx,,cosx) | # # " " = (cosx)(cosx) - (sinx)(-sinx) # # " " = cos^2 x + sin^2 x# # " " = 1 # So we form the two particular solution function: # v_1 = -int \ (p(x)y_2)/(W[y_1,y_2]) \ dx # # \ \ \ = -int \ (cotx sinx)/1 \ dx # # \ \ \ = -int \ cosx/sinx sinx \ dx # # \ \ \ = -int \ cosx \ dx # # \ \ \ = -sinx # And; # v_2 = \ \ \ \ \ int \ (p(x)y_1)/(W[y_1,y_2]) \ dx # # \ \ \ = int \ (cotx cosx )/1\ dx # # \ \ \ = int \ cosx/sinx cosx \ dx # # \ \ \ = int \ cos^2x/sinx \ dx # # \ \ \ = int \ (1-sin^2x)/sinx \ dx # # \ \ \ = int \ cscx-sinx \ dx # # \ \ \ = -ln|cscx+cotx|+cosx # And so we form the Particular solution: # y_p = v_1y_1 + v_2y_2 # # \ \ \ = (-sinx)(cosx) + (cosx-ln|cscx+cotx|)sinx# Which then leads to the GS of [A} # y(x) = y_c + y_p # # \ \ \ \ \ \ \ = Acosx + Bsinx -sinxcosx + sinxcosx-sinxln|cscx+cotx| # # \ \ \ \ \ \ \ = Acosx + Bsinx -sinxln|cscx+cotx| #

Christopher Allers · Answer

undefined The general solution of the given differential equation is:

$$ y(x) = -\sin(x) \ln \left| \csc(x) + \cot(x) \right| + C $$

where $ C $ is an arbitrary constant.