Solve the differential equation by using CF and PI : #(D^2-D-2)y=cos 2x# ?

1 Answer
Steve M
Jun 4, 2018

# y(x) = Ae^(-x)+Be(2x) -3/20cos2x-1/20sin2x #

Explanation:

We have:

# (D^2-D-2)y = cos2x # ..... [A]

Writing in standard form:

# y'' - y' - 2y = cos2x #

This is a second order non-Homogeneous Differentiation Equation. 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.

Complementary Function

The homogeneous equation associated with [A] is

# y'' - y' - 2y = 0 #

And it's associated Auxiliary equation is:

# m^2 -m - 2 = 0 => (m+1)(m-2) = 0#

Which has two real and distinct solutions #m = -1,2#

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 solution of the homogeneous equation [A] is:

# y = Ae^(-x)+Be(2x) #

Particular Solution

In order to find a particular solution of the non-homogeneous equation:

# y'' - y' - 2y = f(x) \ \ # with #f(x) = cos2x #

So, we should probably look for a solution of the form:

# y = acos2x+bsin2x # ..... [B]

Where the constants #a,b# are to be determined by direct substitution and comparison:

Differentiating [B] wrt #x# twice we get:

# y^((1)) = -2asin2x+2bcos2x #
# y^((2)) = -4acos2x-4bsin2x #

Substituting these results into the DE [A] we get:

# (-4acos2x-4bsin2x) - (-2asin2x+2bcos2x) - 2(acos2x+bsin2x) = cos2x #

# :. -4acos2x - 4bsin2x + 2asin2x-2bcos2x - 2acos2x-2bsin2x = cos2x #

# :. (-4a-2b- 2a)cos2x + (-4b+2a-2b)sin2x = cos2x #

# :. (-6a-2b)cos2x + (-6b+2a)sin2x cos2x = cos2x #

Equating coefficients we get:

# cos2x: -6a-2b = 1 #
# sin2x: 6b+2a=0 #

And solving these equations simultaneously, we obtain:

# a=-3/20# and #b=-1/20#

And so we form the Particular solution:

# y_p = -3/20cos2x-1/20sin2x # ..... [B]

General Solution

Which then leads to the GS of [A}

# y(x) = y_c + y_p #
# \ \ \ \ \ \ \ = Ae^(-x)+Be(2x) -3/20cos2x-1/20sin2x #

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

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