How can I solve this differential equation, #y'''-3y''+7y'-5y=x#, using undetermined coefficients?

1 Answer
NJ
Apr 2, 2018

#=>y(x) = [c_1 +c_2cos(2x)+c_3sin(2x)]e^x -1/5x - 7/25#

Explanation:

We are given the differential equation:

#y''' - 3y'' + 7y' - 5y = x#

We know that the solution is of the form:

#y(x)= y_h(x) + y_p(x)#

where #y_h(x)# is the homogenous solution and #y_p(x)# is a particular solution.

================Homogeneous Solution===============

First, we need to solve the homogenous equation:

#y''' - 3y'' + 7y' - 5y = 0#

The characteristic equation is:

#r^3 - 3r^2 + 7r -5 = 0#

where the power of each term corresponds the power of the derivative in the homogeneous equation.

Solving this equation gives the roots:

  • #r = 1#
  • #r = 1-2i#
  • #r = 1+2i#

We have one real root #r = 1#, which will be an exponential term in our solution. The complex roots #r = 1+-2i# correspond to sine and cosine multiplied by exponentials.

This means the homogenous solution is:

#=>color(green)(y_h(x) = c_1e^x +c_2e^xcos(2x)+c_3e^xsin(2x))#

==================Particular Solution=================

Now, we need to find a particular solution. We consider the RHS of the differential equation.

#x# is a first-order polynomial function. So, we will make a guess that our particular solution is also first-order, specifically of the form:

#y_p(x) = alpha x + beta#

where #alpha# and #beta# are undetermined coefficients.

We should calculate up to the third-order derivative of this, since we need to substitute into our differential equation.

  • #y = alpha x + beta#
  • #y' = alpha#
  • #y'' = 0#
  • #y''' = 0#

Substituting into our differential equation, we get:

#y''' - 3y'' + 7y' - 5y = x#

#(0) - 3(0) + 7(alpha) - 5(alphax + beta) = x#

#7alpha - 5alphax - 5 beta = x#

Rearranging to make terms on LHS and RHS align more obviously:

#-5alpha x + (7alpha - 5beta) = x#

We can see immediately that:

  • (1) #-5alpha = 1#
  • (2) #7alpha - 5 beta = 0#

From (1) we find #alpha = -1/5#. Substituting this #alpha# into (2) we find #beta = -7/25#.

Hence, our particular solution is:

#=>color(green)(y_p(x) = -1/5 x -7/25)#

So our final solution is:

#y(x) = y_h(x) + y_p(x)#

#y(x) = c_1e^x +c_2e^xcos(2x)+c_3e^xsin(2x) + (-1/5x-7/25)#

#y(x) = c_1e^x +c_2e^xcos(2x)+c_3e^xsin(2x) -1/5x - 7/25#

Minor simplification (factor out an #e^x#):

#=>color(blue)(y(x) = [c_1 +c_2cos(2x)+c_3sin(2x)]e^x -1/5x - 7/25)#

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