What is the general solution of the differential equation? # x^2y'' +3xy'+17y=0 #

Question

Emma Aldridge · Accepted Answer

#y = (C_1 cos(4 log x)+C_2 sin(4 log x))/x#

Assuming that the differential equation reads

#x^2y''+3xy'+17y=0#

proposing a solution with the structure

#y = c x^alpha# and substituting

#x^2 alpha(alpha-1)cx^(alpha-2)+3alphacx^(alpha-1)+17cx^alpha =(alpha(alpha-1)+3alpha+17)c x^alpha = 0#

Solving now

#alpha(alpha-1)+3alpha+17=0# we obtain

#alpha = -1pm 4i# so the solutions are

#y = c_1 x^(-1-4i)+c_2x^(-1+4i)#

but

#x^(4i) = e^(i4 log x)# and

#e^(i4 log x) = cos(4logx)+isin(4logx)#

so we can reduce the solutions to the form

#y = (C_1 cos(4 log x)+C_2 sin(4 log x))/x#

Luke Alvarez · Answer

# y=(Acos(4lnx))/x+(Bsin(4lnx))/x#

We have: # x^2y'' +3xy'+17y=0 # ..... [A] 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 [A] we get: # x^2((d^2y)/(dt^2)-dy/dt)e^(-2t) +3xe^(-t)dy/dt+17y=0 # # :. ((d^2y)/(dt^2)-dy/dt) +3dy/dt+17y=0 # # :. (d^2y)/(dt^2)+2dy/dt+17y=0 # ..... [B] This is now a second order linear homogeneous 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+2m+17 = 0# We can solve this quadratic equation, and we get two complex conjugate roots: # m=-1+-4i# Thus the Homogeneous equation [B] has the solution: # y=e^(-t)(Acos4t+Bsin4t)# Now we initially used a change of variable: # x = e^t => t=lnx # So restoring this change of variable we get: # y=(x^(-1))(Acos(4lnx)+Bsin(4lnx))# # :. y=(Acos(4lnx))/x+(Bsin(4lnx))/x# Which is the General Solution.

Charles Arocha · Answer

undefined The general solution of the differential equation $x^2y'' + 3xy' + 17y = 0$ can be found by assuming a solution of the form $y = x^r$, where $r$ is a constant. Substituting this into the equation gives a characteristic equation, which can be solved to find the roots $r_1$ and $r_2$. The general solution is then given by:

$$y = C_1x^{r_1} + C_2x^{r_2}$$

where $C_1$ and $C_2$ are arbitrary constants, and $r_1$ and $r_2$ are the roots of the characteristic equation.