Solve the Differential Equation # x^2y'' -3x+5y=0 #?

Question

Paisley Johnson · Accepted Answer

# y=sqrt(x)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x#

We have: # x^2y'' -3x+5y=0 # This is a Euler-Cauchy Equation which is typically solved via a change of variable. Consider the substitution: # x = e^t => xe^(-t)=1# Then we have, #dy/dx = e^(-t)dy/dt#, and, #(d^2y)/(dx^2)=((d^2y)/(dt^2)-dy/dt)e^(-2t)# Substituting into the initial DE we get: # x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) -3e^t+5y=0 # # :. (d^2y)/(dt^2)-dy/dt +5y=3e^t # ..... [A] This is now a second order linear Differentiation Equation. The standard approach is to look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e. # m^2-m+5 = 0# We can solve this quadratic equation, and we get two complex solution: # m=1/2(1+-sqrt(19)i) # Thus the Homogeneous equation: # (d^2y)/(dt^2)-dy/dt +5y=0 # has the solution: #y=e^(1/2t)(Acos(1/2sqrt(19)t) + Bcos(1/2sqrt(19)t))# We now seek a Particular Solution of [A], which will be of the form: # y = Ae^t # Differentiating wrt #t# we get: # y' = Ae^t #, and, ># y'' = Ae^t # Substituting into the DE [A} we get: # Ae^t-Ae^t+5Ae^t=3e^t # Equating coefficients we have: # A-A+5A=3 => A=3/5 # Hence, the GS of [A] is the combination of the homogeneous solution and the particular solution, thus: # y=e^(1/2t)(Acos(1/2sqrt(19)t) + Bcos(1/2sqrt(19)t)) +3/5e^t# Now we initially used a change of variable: # x = e^t => t=lnx # So restoring this change of variable we get: # y=e^(1/2lnx)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5e^lnx# # :. y=e^(lnsqrt(x))(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x# # :. y=sqrt(x)(Acos(1/2sqrt(19)lnx) + Bcos(1/2sqrt(19)lnx)) +3/5x# Which is the General Solution

Ezra Adams · Answer

undefined To solve the differential equation $x^2y'' - 3x + 5y = 0$, we can assume a solution of the form $y = x^m$. Substituting this into the differential equation yields a characteristic equation in terms of $m$. Solving this characteristic equation gives us the values of $m$. Depending on the nature of the roots of the characteristic equation, the general solution can have different forms.