Solving Non-Homogeneous Linear DEs with Constant Coefficients, Lecture notes of Differential Equations

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SOLUTIONS OF NON-

HOMOGENEOUS LINEAR

DIFFERENTIAL EQUATIONS

WITH CONSTANT

COEFFICIENTS

INTENDED LEARNING OUTCOME:

• Solve for the solutions of non-homogeneous linear D.E. with

constant coefficients using:

✓ Method of Reduction of Order

✓ Method of Undetermined Coefficients

✓ Method of Variation of Parameters

• Method of Reduction of Order
• Method of Undetermined Coefficients
• Method of Variation of Parameters SOLUTIONS OF NON-HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS WITH CONSTANT COEFFICIENTS

This method may be advantageously applied in finding 𝑌

𝑃

under the following conditions:

• The roots of the auxiliary equation ∅ 𝑚 = 0 are all real.
• The order of ∅ 𝐷 𝑦 = 𝑓 𝑥 is not too large.
• The functional operator ∅ 𝐷 is expressible in the

factored form

0

1

2

𝑛

The solution is finally given by 𝐲 = 𝐘𝐩 = 𝐅(𝐱)

Example : Find the general solution of 𝐷 2 − 4 𝑦 = 4 𝑥 − 3 𝑒 𝑥 ∅ 𝑚 = 𝑚 2 − 4 = 0 ∅ 𝑚 = 𝑚 − 2 𝑚 + 2 = 0 Complementary Function: 𝒀𝒄 = 𝒄𝟏𝒆 𝟐𝒙

• 𝒄𝟐𝒆 −𝟐𝒙 Particular Solution, 𝑌𝑝: 𝐷 − 2 𝐷 + 2 𝑦 = 4 𝑥 − 3 𝑒 𝑥 Let: 𝐷 + 2 𝑦 = 𝑧 𝐷 − 2 𝑧 = 4 𝑥 − 3 𝑒 𝑥 𝑑𝑧 𝑑𝑥

𝑥

Example : Find the general solution of 𝐷 2 − 4 𝑦 = 4 𝑥 − 3 𝑒 𝑥 𝑧𝑒 −2𝑥 = න 𝑒 −2𝑥 4 𝑥 − 3 𝑒 𝑥 𝑑𝑥 𝑧𝑒 −2𝑥 = 4 න^ 𝑥𝑒 −2𝑥 𝑑𝑥 − 3 න^ 𝑒 −𝑥 𝑑𝑥 𝑧𝑒 −2𝑥 = 4 −

−2𝑥 − −

−2𝑥 − 3 −𝑒 −𝑥 𝑧𝑒 −2𝑥 = −2𝑥𝑒 −2𝑥 − 𝑒 −2𝑥

• 3 𝑒 −𝑥 𝑧 = −2𝑥 − 1 + 3 𝑒 𝑥

2 𝑥

Example : Find the general solution of 𝐷 2 − 4 𝑦 = 4 𝑥 − 3 𝑒 𝑥 𝑧 = − 2 𝑥 − 1 + 3 𝑒 𝑥 𝑫 + 𝟐 𝒚 = 𝒛 𝐷 + 2 𝑦 = −2𝑥 − 1 + 3 𝑒 𝑥 𝑃 = 2 𝑄 = − 2 𝑥 − 1 + 3 𝑒 𝑥 ∅ = 𝑒 ׬ 𝑃𝑑𝑥^ = 𝑒 2𝑥 𝑦 ∅ = න ∅ 𝑄 𝑑𝑥 𝑦𝑒 2𝑥 = න 𝑒 2𝑥 −2𝑥 − 1 + 3 𝑒 𝑥 𝑑𝑥

Example : Find the general solution of 𝐷 2 − 4 𝑦 = 4 𝑥 − 3 𝑒 𝑥 𝒀𝒄 = 𝒄𝟏𝒆 𝟐𝒙

• 𝒄𝟐𝒆 −𝟐𝒙 𝒚 = 𝒆 𝒙 − 𝒙 = 𝒀𝒑 The complete solution: 𝒚 = 𝒀𝒑 + 𝒀𝒄 𝒚 = 𝒆 𝒙 − 𝒙 + 𝒄𝟏𝒆 𝟐𝒙
• 𝒄𝟐𝒆 −𝟐𝒙

Example : Find the general solution of 𝐷 3 − 2 𝐷 2 +𝐷 𝑦 = 𝑥 ∅ 𝑚 = 𝑚 3 − 2 𝑚 2 +𝑚 = 0 ∅ 𝑚 = 𝑚 𝑚 2 − 2𝑚 + 1 = 0 ∅ 𝑚 = 𝑚 𝑚 − 1 𝑚 − 1 = 0 Complementary Function: 𝒀𝒄 = 𝒆 𝒙 (𝒄𝟏+𝒄𝟐𝒙) + 𝒄𝟑 Particular Solution, 𝑌𝑝: 𝐷 𝐷 − 1 𝐷 − 1 𝑦 = 𝑥 Let: 𝐷 − 1 𝐷 − 1 𝑦 = 𝑧 𝐷𝑧 = 𝑥 𝑧 =

2

Example : Find the general solution of 𝐷 3 − 2 𝐷 2 +𝐷 𝑦 = 𝑥 𝑑𝑣 𝑑𝑥

2 𝑃 = − 1 𝑄 =

2 ∅ = 𝑒 ׬ 𝑃𝑑𝑥^ = 𝑒 −𝑥 𝑣 ∅ = න ∅ 𝑄 𝑑𝑥 𝑣𝑒 −𝑥 = න 𝑒 −𝑥

2 𝑑𝑥

Example : Find the general solution of 𝐷 3 − 2 𝐷 2 +𝐷 𝑦 = 𝑥 𝑣𝑒 −𝑥 = න 𝑒 −𝑥

2 𝑑𝑥 𝑣𝑒 −𝑥 = 1 2

2 𝑒 −𝑥 𝑑𝑥 𝑣𝑒 −𝑥 =

2 𝑒 −𝑥 − න −𝑒 −𝑥 2𝑥 𝑑𝑥 𝑣𝑒 −𝑥 =

2 𝑒 −𝑥

• 2 −𝑥𝑒 −𝑥 − න −𝑒 −𝑥 𝑑𝑥) 𝑣𝑒 −𝑥 = − 1 2

2 𝑒 −𝑥 −𝑥𝑒 −𝑥

• (−𝑒 −𝑥

Example : Find the general solution of 𝐷 3 − 2 𝐷 2 +𝐷 𝑦 = 𝑥 𝐷 − 1 𝑦 = −

2 − 𝑥 − 1 𝑃 = − 1 𝑄 = −

2 − 𝑥 − 1 ∅ = 𝑒 ׬ 𝑃𝑑𝑥^ = 𝑒 −𝑥 𝑓𝑟𝑜𝑚 𝒚 ∅ = (^) ׬ ∅ 𝑸 𝒅𝒙; 𝑦𝑒 −𝑥 = (^) ׬ 𝑒 −𝑥 − 1 2

2 − 𝑥 − 1 𝑑𝑥 𝑦𝑒 −𝑥 = −

2 𝑒 −𝑥 𝑑𝑥 − න 𝑥𝑒 −𝑥 𝑑𝑥 − න 𝑒 −𝑥 dx 𝑦𝑒 −𝑥 = − −

2 𝑒 −𝑥 − 𝑥𝑒 −𝑥

• (−𝑒 −𝑥

−𝑥

• (−𝑒 −𝑥

−𝑥

Example : Find the general solution of 𝐷 3 − 2 𝐷 2 +𝐷 𝑦 = 𝑥 𝑦𝑒 −𝑥 = න 𝑒 −𝑥 −

2 − 𝑥 − 1 𝑑𝑥 𝑦𝑒 −𝑥 = −

2 𝑒 −𝑥 𝑑𝑥 − න^ 𝑥𝑒 −𝑥 𝑑𝑥 − න^ 𝑒 −𝑥 dx 𝑦𝑒 −𝑥 = − −

2 𝑒 −𝑥 − 𝑥𝑒 −𝑥

• (−𝑒 −𝑥

−𝑥

• (−𝑒 −𝑥

−𝑥 𝑦𝑒 −𝑥 =

2 𝑒 −𝑥

• 2 𝑥𝑒 −𝑥
• 3 𝑒 −𝑥 𝒚 =

𝟐

• 𝟐𝒙 + 𝟑 = 𝒀𝒑

[ ] 𝑒

𝑥