Solve the differential equation # (D^2+2D+5)y=xe^x # ?

Question

Christopher Agresta · Accepted Answer

# y = Ae^(-x)cos2x + Be^(-x)sin2x + (xe^x)/8 - e^x/16#

We have: # (D^2+2D+5)y=xe^x # Where #D# is the linear differential operator #d/dx#. Thus we can write the equation as: # (d^2y)/(dx^2) + 2dy/dx + 5y = xe^x # ..... [A] This is a second 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 quadratic equation with the coefficients of the derivatives, and then finding an independent particular solution, #y_p# of the non-homogeneous equation. Complimentary Function The homogeneous equation associated with [A] is # (d^2y)/(dx^2) + 2dy/dx + 5y = 0# And it's associated Auxiliary equation is: # m^2 + 2m+5 = 0 # Which has two complex solutions #m=-1+-2i# Thus the solution of the homogeneous equation is: # y_c = e^(-1x)(Acos2x + Bsin2x) # # \ \ \ = Ae^(-x)cos2x + Be^(-x)sin2x # Particular Solution In order to find a particular solution of the non-homogeneous equation we would look for a solution of the form: # y = (ax+b)e^x # Where the constants #a# and #b# are to be determined by direct substitution and comparison: Differentiating wrt #x# (using the product rule) we get: # dy/dx = (ax+b)e^x+(a)e^x # # " " = (ax+a+b)e^x # Differentiating again wrt #x# (using the product rule) we get: # (d^2y)/(dx^2) = (ax+a+b)e^x+(a)e^x # # " " = (ax+2a+b)e^x # Substituting into the DE [A] we get: # (ax+2a+b)e^x + 2(ax+a+b)e^x +5(ax+b)e^x = xe^x# # :. (ax+2a+b) + 2(ax+a+b) +5(ax+b) = x# Equating coefficients of #x# and constants we get # x^1: (2a+b) + 2(a+b) +5(b) = 0 => 4a+8b=0# # x^0: (a) + 2(a) +5(a) = 1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \=> 8a=1# Solving simultaneously, we have: #a=1/8# #1/2+8b=0=>b=-1/16# And so we form the Particular solution: # y_p = (1/8x-1/6)e^x # # \ \ \ = (xe^x)/8 - e^x/16# Which then leads to the GS of [A} # y(x) = y_c + y_p # # \ \ \ \ \ \ \ = Ae^(-x)cos2x + Be^(-x)sin2x + (xe^x)/8 - e^x/16#

Ezekiel Alexander · Answer

undefined To solve the differential equation $(D^2 + 2D + 5)y = xe^x$, where $D$ represents the differential operator with respect to $x$, we can follow these steps:

1. Find the complementary function $y_c$ by solving the homogeneous equation $(D^2 + 2D + 5)y_c = 0$.
2. Find a particular solution $y_p$ for the non-homogeneous equation $(D^2 + 2D + 5)y_p = xe^x$.
3. The general solution is given by $y = y_c + y_p$.

First, solve the homogeneous equation:

$(D^2 + 2D + 5)y_c = 0$

The characteristic equation is $m^2 + 2m + 5 = 0$, which has complex roots $m = -1 \pm 2i$.

So, the complementary function is $y_c = e^{-x}(A\cos(2x) + B\sin(2x))$, where $A$ and $B$ are arbitrary constants.

Next, find the particular solution $y_p$ for the non-homogeneous equation. Since $xe^x$ is a product of a polynomial and an exponential function, we can try the method of undetermined coefficients. We assume $y_p$ takes the form $y_p = (Ax + B)e^x$.

Now, differentiate $y_p$ twice and substitute into the differential equation to find $A$ and $B$.

$(D^2 + 2D + 5)y_p = (A + 2A + 5Ax + 2B)e^x = xe^x$

Comparing coefficients, we get $A = \frac{-1}{5}$ and $B = \frac{1}{25}$.

Therefore, the particular solution is $y_p = \frac{-1}{5}xe^x + \frac{1}{25}e^x$.

The general solution is $y = y_c + y_p = e^{-x}(A\cos(2x) + B\sin(2x)) + \frac{-1}{5}xe^x + \frac{1}{25}e^x$, where $A$ and $B$ are arbitrary constants.