What is the general solution of the differential equation # y'''-y''+44y'-4=0 #?

Question

Isabella Ackerman · Accepted Answer

# y = A + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)} + x#

We have: # y'''-y''+44y'-4=0 # Alternatively, as an alternative: # y'''-y''+4y' = 4 # ..... [A] This is a third 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 polynomial 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'''-y''+4y' = 0 # ..... [B] Additionally, the related auxiliary equation is: # m^3-m^2+4m = 0# The challenge with higher order Differential Equation is solving the associated higher order Auxiliary equation. By inspection we see #m# is a factor, so we get:: # m(m^2-m+4) = 0# We can also finish the square by: # m((m-1/2)^2-(1/2)^2+4) = 0# # :. m((m-1/2)^2+4-1/4) = 0# # :. m((m-1/2)^2+15/4) = 0# Consequently, we have the answers: # m=0 # # (m-1/2)^2 = -15/4 => m = 1/2+sqrt(15)/2-# Parts of the solution are determined by the auxiliary equation's roots; if these parts are linearly independent, the solutions' superposition forms the complete general solution. Thus, the homogeneous equation [B] has the following solution: # y = Ae^(0x) + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)}# # \ \ = A + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)}# Specific Resolution To identify a specific non-homogeneous equation solution: # y'''-y''+4y' = f(x) \ \ # with #f(x)=4 # ..... [C] then as #f(x)# is a polynomial of degree #0#, we would look for a polynomial solution of the same degree, i.e. of the form #y = a# However, such a solution already exists in the CF solution and so must consider a potential solution of the form #y=ax#, Where the constants #a# is to be determined by direct substitution and comparison: Differentiating #y=ax# wrt #x# we get: # y' = a # # y'' = 0 # # y''' = 0 # When we enter these outcomes into the DE [A], we obtain: # 0-0+4a = 4 => a=1 # Thus, we arrive at the specific solution: # y_p = x # Overall Resolution which ultimately results in the GS of [A} # y(x) = y_c + y_p # # \ \ \ \ \ \ \ = A + e^(1/2x){Bcos(sqrt(15)/2x) + Csin(sqrt(15)/2x)} + x# Note this solution has #3# constants of integration and #3# linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution

Christopher Allers · Answer

#"Characteristic equation is : "#
#z^3 - z^2 + 4 z = 0#
#=> z(z^2 - z + 4) = 0#
#=> z = 0  " OR " z^2 - z + 4 = 0#
#"disc of the quad. eq. = 1 - 16 = -15 < 0"#
#"so we have two complex solutions, they are"#
#z = (1 pm sqrt(15) i)/2#
#"So the general solution of the homogeneous equation is : "#
#A + B' exp(x/2) exp((sqrt(15)/2) i x) +#
#C' exp(x/2) exp(-(sqrt(15)/2) i x)#
#= A + B exp(x/2) cos(sqrt(15)x /2) + C exp(x/2) sin(sqrt(15) x/2)#
#"The particular solution to the complete equation is"#
#"y=x ,"#
#"That is easy to see."#
#"So the complete solution is :"#
#y(x) = x + A + B exp(x/2) cos(sqrt(15) x/2) + C exp(x/2) sin(sqrt(15) x/2)#

Luke Alvarez · Answer

undefined The general solution of the differential equation $y''' - y'' + 44y' - 4 = 0$ is:

$$y(t) = C_1 e^{2t} + C_2 e^{-2t} + C_3 e^{-22t} + \frac{1}{11}$$

where $C_1$, $C_2$, and $C_3$ are arbitrary constants.