Solve the differential equation by using CF and PI : (D^2-D-2)y=cos 2x(D2−D−2)y=cos2x ?

1 Answer
Steve M
Jun 4, 2018

y(x) = Ae^(-x)+Be(2x) -3/20cos2x-1/20sin2x y(x)=Ae−x+Be(2x)−320cos2x−120sin2x

Explanation:

We have:

(D^2-D-2)y = cos2x (D2−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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