Question

Solve the given differential equation by undetermined coefficients.

\(y''' - 3y'' + 3y' - y = x - 4{e^x}\)

Short answer

The general solution of the given differential equation is

\(y = {c_1}{e^x} + {c_2}x{e^x} + {c_3}{x^2}{e^{3x}} - \frac{2}{3}{x^3}{e^x} - x - 3\)

Step-by-step solution

Note the given data

Given the third order differential equation \(y''' - 3y'' + 3y' - y = x - 4{e^x}\)

Obtain the general solution of the homogeneous differential equation  

Weknow ahomogeneous differential equation is an equation containing

differentiation and a function with a set of variables.

First, we need to solve the associated homogeneous equation\(y''' - 3y'' + 3y' - y = 0\)

Corresponding auxiliary equation is\({m^3} - 3{m^2} + 3m - 1 = 0\)

\(({m^2} - 2m + 1)(m - 1) = 0\)

\((m - 1)(m - 1)(m + 2) = 0\)

Solving,

m₁ = 1, m₂ =1 and m₃ = -2, which are real and repeated.

So the solution of corresponding homogeneous equation is

\({y_h} = {c_1}{e^x} + {c_2}x{e^x} + {c_3}{x^2}{e^{3x}}\)

Finding the particular solution

Given non homogeneous differential equation is\(y''' - 3y'' + 3y' - y = x - 4{e^x}\)

Assume that \({y_p} = A{x^3}{e^x} + Bx + C\) …..(1)is a solution of the non-homogeneous differential equation

Differentiate with respect to x

\({y_p}' = A{x^3}{e^x} + 3A{x^2}{e^x} + B\)……(2)

\({y_p}'' = A{x^3}{e^x} + 6A{x^2}{e^x} + 6Ax{e^x}\)…..(3)

\({y_p}''' = A{x^3}{e^x} + 9A{x^2}{e^x} + 18Ax{e^x} + 6A{e^x}\)……(4)

Substituting (1) ,(2), (3) and (4) in given differential equation

\(\begin{aligned}{l}A{x^3}{e^x} + 9A{x^2}{e^x} + 18Ax{e^x} + 6A{e^x} - 3[A{x^3}{e^x} + 6A{x^2}{e^x} + 6Ax{e^x}] + 3[A{x^3}{e^x} + 3A{x^2}{e^x} + B]\\ - [A{x^3}{e^x} + Bx + C] = x - 4{e^x}\end{aligned}\)

Or, \(Bx + (3B + C) + 6A{e^x} = x - 4{e^x}\)

comparing the coefficients,

\( - B = 1\)

\(3B - C = 0\)

\(6A = - 4\)

Solving, \(A = \frac{{ - 2}}{3},B = - 1,C = - 3\)

Hence the particular solution is

\({y_p} = \frac{{ - 2}}{3}{x^3}{e^x} - x - 3\)

Finding the solution

Hence the solution of given differential equation is

y=y_h+y_p

\(y = {c_1}{e^x} + {c_2}x{e^x} + {c_3}{x^2}{e^{3x}} - \frac{2}{3}{x^3}{e^x} - x - 3\)