Question: Find the particular solution #y_p(x)# of homogeneous linear equation of the 2. order with constant coefficients. How do I solve it? Thank you! (pictures below)

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this should be solution:
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2 Answers
Andrea S.
May 18, 2018

Solve the characteristic equation:

#lambda^2 -10lambda+25 = 0#

#(lambda-5)^2 =0#

so we have a single root with double multiplicity:

#lambda = 5#

In such case the general solution of the equation is:

#y(x) = c_1e^(5x)+c_2xe^(5x)#

Evaluate the first derivative:

#y'(x) = 5c_1e^(5x) + c_2e^(5x) + 5c_2xe^(5x)#

and determine the constants from the initial conditions:

#{(y(1) = e^5),(y'(1) = 4e^5):}#

#{(c_1e^5 +c_2e^5 = e^5),(5c_1e^5 +c_2e^5 +5c_2e^5 = 4e^5):}#

#{(c_1 +c_2 = 1),(5c_1 + 6c_2= 4):}#

#{(c_1 = 2),(c_2= -1):}#

Then the required solution is:

#y(x) = (2-x)e^(5x)#

Steve M
May 18, 2018

# y = (2-x)e^(5x) #

Explanation:

We seek a solution to

# y'' + 10y' +25y = 0# with #y'(1)=4e^5, y(1)=e^5# ..... [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 polynomial equation with the coefficients of the derivatives., and then finding an independent particular solution, #y_p# of the non-homogeneous equation.

Complementary Function

The homogeneous equation associated with [A] is

# y'' + 10y' +25y = 0 # ..... [B]

And it's associated Auxiliary equation is:

# m^2 +10m +25 = 0 => (m+5)^2=0#

And so we have the solutions:

# m = +-5 \ \ \ \ # (repeated real)

The roots of the auxiliary equation determine parts of the solution, which if linearly independent then the superposition of the solutions form the full general solution.

  • Real distinct roots #m=alpha,beta, ...# will yield linearly independent solutions of the form #y_1=Ae^(alphax)#, #y_2=Be^(betax)#, ...
  • Real repeated roots #m=alpha#, will yield a solution of the form #y=(Ax+B)e^(alphax)# where the polynomial has the same degree as the repeat.
  • Complex roots (which must occur as conjugate pairs) #m=p+-qi# will yield a pairs linearly independent solutions of the form #y=e^(px)(Acos(qx)+Bsin(qx))#

Thus the General solution of the homogeneous equation [B] is:

# y = (Ax+B)e^(5x) #

Note this solution has #2# constants of integration and #2# linearly independent solutions, hence by the Existence and Uniqueness Theorem their superposition is the General Solution

Now, we form the first derivative using the product rule

# y' = (Ax+B)(5e^(5x)) + (A)(e^(5x)) #
# \ \ \ \= (5Ax+5B+A)e^(5x) #

and applying the initial conditions:

# {: (y(1)=e^5, => (A+B)e^(5) = 4^5, => A+B =1), (y'(1)=4e^5, => (5A+5B+A)e^(5) = 4e^5, => 6A+5B=4 ) :} #

And solving simultaneously we get the solution:

# A=-1# and #B=2 #

Thus we the the Particular Solution:

# y = (-x+2)e^(5x) #

# :. y = (2-x)e^(5x) \ \ \ # QED

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