Non Homogeneous Equations-method Of Undetermined Coefficients

Ch 3.6: Nonhomogeneous Equations;     Method of Undetermined Coefficients Recall the nonhomogeneous equation y′′ + p (t ) y′ + q (t ) y = g (t ) where p, q, g are continuous functions on an open interval I. The associated homogeneous equation is y′′ + p (t ) y′ + q (t ) y = 0 In this section we will learn the method of undetermined coefficients to solve the nonhomogeneous equation, which relies on knowing solutions to homogeneous equation.

Theorem 3.6.1 If Y1, Y2 are solutions of nonhomogeneous equation y′′ + p (t ) y′ + q(t ) y = g (t ) then Y1 - Y2 is a solution of the homogeneous equation y′′ + p (t ) y′ + q (t ) y = 0 If y1, y2 form a fundamental solution set of homogeneous equation, then there exists constants c1, c2 such that Y1 (t ) − Y2 (t ) = c1 y1 (t ) + c2 y2 (t )

Theorem 3.6.2 (General Solution) The general solution of nonhomogeneous equation y′′ + p (t ) y′ + q(t ) y = g (t ) can be written in the form y (t ) = c1 y1 (t ) + c2 y2 (t ) + Y (t ) where y1, y2 form a fundamental solution set of homogeneous equation, c1, c2 are arbitrary constants and Y is a specific solution to the nonhomogeneous equation.

Method of Undetermined Coefficients Recall the nonhomogeneous equation y′′ + p (t ) y′ + q(t ) y = g (t ) with general solution y (t ) = c1 y1 (t ) + c2 y2 (t ) + Y (t ) In this section we use the method of undetermined coefficients to find a particular solution Y to the nonhomogeneous equation, assuming we can find solutions y1, y2 for the homogeneous case. The method of undetermined coefficients is usually limited to when p and q are constant, and g(t) is a polynomial, exponential, sine or cosine function.

Example 1: Exponential g(t) Consider the nonhomogeneous equation y′′ − 3 y′ − 4 y = 3e 2t We seek Y satisfying this equation. Since exponentials replicate through differentiation, a good start for Y is: Y (t ) = Ae 2t ⇒ Y ′(t ) = 2 Ae 2t , Y ′′(t ) = 4 Ae 2t

Substituting these derivatives into differential equation, 4 Ae 2t − 6 Ae 2t − 4 Ae 2t = 3e 2t ⇔ − 6 Ae 2t = 3e 2t

⇔ A = −1 / 2

Thus a particular solution to the nonhomogeneous ODE is 1 2t Y (t ) = − e 2

Example 2: Sine g(t), First Attempt

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Consider the nonhomogeneous equation y′′ − 3 y′ − 4 y = 2 sin t We seek Y satisfying this equation. Since sines replicate through differentiation, a good start for Y is: Y (t ) = A sin t ⇒ Y ′(t ) = A cos t , Y ′′(t ) = − A sin t

Substituting these derivatives into differential equation, − A sin t − 3 A cos t − 4 A sin t = 2 sin t ⇔ ( 2 + 5 A) sin t + 3 A cos t = 0 ⇔ c1 sin t + c2 cos t = 0 Since sin(x) and cos(x) are linearly independent (they are not multiples of each other), we must have c1= c2 = 0, and hence 2 + 5A = 3A = 0, which is impossible.

y′′ − 3 y′ − 4 y = 2 sin t

Example 2: Sine g(t), Particular Solution

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Our next attempt at finding a Y is Y (t ) = A sin t + B cos t ⇒ Y ′(t ) = A cos t − B sin t , Y ′′(t ) = − A sin t − B cos t

Substituting these derivatives into ODE, we obtain

( − A sin t − B cos t ) − 3( A cos t − B sin t ) − 4( A sin t + B cos t ) = 2 sin t ⇔ ( − 5 A + 3B ) sin t + ( − 3 A − 5 B ) cos t = 2 sin t ⇔ − 5 A + 3B = 2, − 3 A − 5 B = 0 ⇔ A = −5 / 17, B = 3 / 17

Thus a particular solution to the nonhomogeneous ODE is −5 3 Y (t ) = sin t + cos t 17 17

Example 3: Polynomial g(t) Consider the nonhomogeneous equation y′′ − 3 y′ − 4 y = 4t 2 − 1 We seek Y satisfying this equation. We begin with Y (t ) = At 2 + Bt + C ⇒ Y ′(t ) = 2 At + B, Y ′′(t ) = 2 A

Substituting these derivatives into differential equation, 2 A − 3( 2 At + B ) − 4( At 2 + Bt + C ) = 4t 2 − 1

⇔ − 4 At 2 − ( 6 A + 4 B ) t + ( 2 A − 3B − 4C ) = 4t 2 − 1 ⇔ − 4 A = 4, 6 A + 4 B = 0, 2 A − 3B − 4C = −1 ⇔ A = −1, B = 3 / 2 , C = −11 / 8

Thus a particular solution to the nonhomogeneous ODE is 3 11 Y (t ) = − t 2 + t − 2 8

Example 4: Product g(t) Consider the nonhomogeneous equation y′′ − 3 y′ − 4 y = −8e t cos 2t

We seek Y satisfying this equation, as follows: Y (t ) = Aet cos 2t + Bet sin 2t Y ′(t ) = Aet cos 2t − 2 Aet sin 2t + Be t sin 2t + 2 Bet cos 2t = ( A + 2 B ) e t cos 2t + ( − 2 A + B ) et sin 2t

Y ′′(t ) = ( A + 2 B ) e t cos 2t − 2( A + 2 B ) et sin 2t + ( − 2 A + B ) et sin 2t + 2( − 2 A + B ) et cos 2t

= ( − 3 A + 4 B ) et cos 2t + ( − 4 A − 3 B ) et sin 2t

Substituting derivatives into ODE and solving for A and B: 10 2 10 t 2 t A= , B= ⇒ Y (t ) = e cos 2t + e sin 2t 13 13 13 13

Discussion: Sum g(t) Consider again our general nonhomogeneous equation y′′ + p(t ) y′ + q(t ) y = g (t ) Suppose that g(t) is sum of functions: g (t ) = g1 (t ) + g 2 (t )

If Y1, Y2 are solutions of

y′′ + p (t ) y′ + q (t ) y = g1 (t ) y′′ + p (t ) y′ + q (t ) y = g 2 (t ) respectively, then Y1 + Y2 is a solution of the nonhomogeneous equation above.

Example 5: Sum g(t) Consider the equation y′′ − 3 y′ − 4 y = 3e 2t + 2 sin t − 8et cos 2t Our equations to solve individually are y′′ − 3 y′ − 4 y = 3e 2t y′′ − 3 y′ − 4 y = 2 sin t y′′ − 3 y′ − 4 y = −8e t cos 2t

Our particular solution is then 1 3 5 10 2 Y (t ) = − e 2t + cos t − sin t + e t cos 2t + e t sin 2t 2 17 17 13 13

Example 6: First Attempt

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Consider the equation y′′ + 4 y = 3 cos 2t We seek Y satisfying this equation. We begin with Y (t ) = A sin 2t + B cos 2t ⇒ Y ′(t ) = 2 A cos 2t − 2 B sin 2t , Y ′′(t ) = −4 A sin 2t − 4 B cos 2t

Substituting these derivatives into ODE:

( − 4 A sin 2t − 4 B cos 2t ) + 4( A sin 2t + B cos 2t ) = 3 cos 2t ( − 4 A + 4 A) sin 2t + ( − 4 B + 4 B ) cos 2t = 3 cos 2t 0 = 3 cos 2t

Thus no particular solution exists of the form Y (t ) = A sin 2t + B cos 2t

Example 6: Homogeneous Solution

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Thus no particular solution exists of the form Y (t ) = A sin 2t + B cos 2t

To help understand why, recall that we found the corresponding homogeneous solution in Section 3.4 notes: y′′ + 4 y = 0 ⇒ y (t ) = c1 cos 2t + c2 sin 2t Thus our assumed particular solution solves homogeneous equation y′′ + 4 y = 0 instead of the nonhomogeneous equation. y′′ + 4 y = 3 cos 2t

y′′ + 4 y = 3 cos 2t

Example 6: Particular Solution

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Our next attempt at finding a Y is: Y (t ) = At sin 2t + Bt cos 2t Y ′(t ) = A sin 2t + 2 At cos 2t + B cos 2t − 2 Bt sin 2t Y ′′(t ) = 2 A cos 2t + 2 A cos 2t − 4 At sin 2t − 2 B sin 2t − 2 B sin 2t − 4 Bt cos 2t = 4 A cos 2t − 4 B sin 2t − 4 At sin 2t − 4 Bt cos 2t

Substituting derivatives into ODE, 4 A cos 2t − 4 B sin 2t = 3 cos 2t ⇒ A = 3 / 4, B = 0 3 ⇒ Y (t ) = t sin 2t 4