TLDR;
This video serves as an introductory guide to solving higher-order non-homogeneous differential equations with constant coefficients, a key topic in Engineering Mathematics-IV (M4) under the C-24 pattern. It explains the fundamental concepts, highlights the differences between homogeneous and non-homogeneous equations, and outlines the methods for finding solutions using complementary functions and particular integrals.
- Introduces higher-order non-homogeneous differential equations.
- Explains the difference between homogeneous and non-homogeneous equations.
- Outlines methods for finding solutions using complementary functions and particular integrals.
Introduction to Higher Order Non-Homogeneous Equations [0:02]
The video introduces the topic of higher-order non-homogeneous differential equations with constant coefficients, a key area in Engineering Mathematics-IV (M4) under the C-24 pattern. It builds upon the knowledge of homogeneous differential equations from the second unit, explaining that the methods used for solving homogeneous equations will be applied here. The presenter encourages viewers to subscribe, like, comment, and follow on Instagram and Telegram for more content.
Defining Non-Homogeneous Differential Equations [1:06]
The presenter defines a higher-order non-homogeneous equation as one in the form aₙ(dⁿy/dxⁿ) + a₁(dⁿ⁻¹y/dxⁿ⁻¹) + ... + a₀y = X, where X is a function of x. If X is zero, the equation is homogeneous; otherwise, it's non-homogeneous. In this unit, X can take forms like e^(ax), sin(bx), cos(bx), or polynomial expressions. The presenter also reminds the audience that d/dx is represented as 'D', allowing the equation to be written as f(D)y = X, which defines a non-homogeneous differential equation. If f(D)y = 0, it is a homogeneous differential equation, a concept already covered in the second unit.
Methods to Solve Non-Homogeneous Equations [3:54]
The video outlines that solving f(D)y = X involves finding both the complementary function and the particular integral. The complementary function is similar to what was learned in the second unit, while the particular integral is a new concept. The general solution is the sum of these two. The presenter introduces the formula for the particular integral as 1/f(D) * X. For a quadratic equation like aD² + bD + c, the solution involves factorisation to find two roots. The presenter then introduces the first case where f(D)y = e^(ax), explaining that the particular integral is always 1/f(D) * X.
Case 1: Solving Equations with e^(ax) [5:44]
The presenter explains the first method for solving non-homogeneous differential equations when f(D)y = e^(ax). The particular integral is given by 1/f(D) * e^(ax). The presenter introduces an alternative formula for the particular integral: e^(ax) * integration of e^(-ax) * X dx, but suggests a simpler method. The presenter advises substituting D = a into the equation. If f(a) ≠ 0, the particular integral is simply 1/f(a) * e^(ax). However, if f(a) = 0, the solution changes based on the degree of the term.
Adjusting Solutions When f(a) = 0 [7:27]
If substituting D = a results in f(a) = 0, the solution is adjusted based on the degree of the term. For 1/(D-a) * e^(ax), the solution is x/1! * e^(ax). For 1/(D-a)² * e^(ax), the solution is x²/2! * e^(ax). Generally, for 1/(D-a)ⁿ * e^(ax), the solution is xⁿ/n! * e^(ax). The presenter provides a general formula for these cases. The presenter also introduces formulas for hyperbolic functions: sinh(x) = (e^(ax) - e^(-ax))/2 and cosh(x) = (e^(ax) + e^(-ax))/2.
Solving Example 1: d²y/dx² - 4dy/dx + 3y = e^(2x) [10:14]
The presenter begins solving the differential equation d²y/dx² - 4dy/dx + 3y = e^(2x). The presenter identifies that the solution requires finding both the complementary function and the particular integral. The given differential equation is in the form f(D)y = X, where f(D) = D² - 4D + 3 and X = e^(2x). The auxiliary equation is m² - 4m + 3 = 0, which factors to (m-1)(m-3) = 0, giving roots m = 1 and m = 3. Since the roots are real and distinct, the complementary function is C₁e^(x) + C₂e^(3x).
Finding the Particular Integral for Example 1 [14:23]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X, which is 1/(D² - 4D + 3) * e^(2x). The presenter substitutes D = 2 into the factorised form (D-1)(D-3), resulting in (2-1)(2-3) = -1. Therefore, the particular integral is -e^(2x). The general solution is the sum of the complementary function and the particular integral: y = C₁e^(x) + C₂e^(3x) - e^(2x).
Solving Example 2: 4d²y/dx² + 4dy/dx - 3y = e^(2x) [17:18]
The presenter solves the differential equation 4d²y/dx² + 4dy/dx - 3y = e^(2x). The equation is in the form f(D)y = X, where f(D) = 4D² + 4D - 3 and X = e^(2x). The auxiliary equation is 4m² + 4m - 3 = 0, which factors to (2m+3)(2m-1) = 0, giving roots m = -3/2 and m = 1/2. Since the roots are real and distinct, the complementary function is C₁e^(-3x/2) + C₂e^(x/2).
Finding the Particular Integral for Example 2 [21:48]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X, which is 1/(4D² + 4D - 3) * e^(2x). The presenter substitutes D = 2 into the factorised form (2D+3)(2D-1), resulting in (22+3)(22-1) = 7 * 3 = 21. Therefore, the particular integral is e^(2x)/21. The general solution is the sum of the complementary function and the particular integral: y = C₁e^(-3x/2) + C₂e^(x/2) + e^(2x)/21.
Example 3: d²y/dx² + dy/dx + y = e^(4x) and Complex Roots [24:51]
The presenter solves the differential equation d²y/dx² + dy/dx + y = e^(4x). The equation is rewritten as (D² + D + 1)y = e^(4x), where f(D) = D² + D + 1 and X = e^(4x). The auxiliary equation is m² + m + 1 = 0. Since factorisation is not possible, the quadratic formula is used to find the roots: m = (-1 ± √(-3))/2 = -1/2 ± (√3/2)i. The roots are complex, in the form α ± iβ, where α = -1/2 and β = √3/2.
Finding the Particular Integral and General Solution for Example 3 [27:06]
The presenter finds the complementary function and particular integral for the equation. Since the roots are complex, the complementary function is e^(αx)(C₁cos(βx) + C₂sin(βx)) = e^(-x/2)(C₁cos((√3/2)x) + C₂sin((√3/2)x)). The particular integral is 1/f(D) * X = 1/(D² + D + 1) * e^(4x). Substituting D = 4, the particular integral becomes e^(4x)/(4² + 4 + 1) = e^(4x)/21. The general solution is the sum of the complementary function and the particular integral: y = e^(-x/2)(C₁cos((√3/2)x) + C₂sin((√3/2)x)) + e^(4x)/21.
Example 4: d²y/dx² + 5dy/dx + 6y = e^(-2x) and Handling Zero Denominators [30:06]
The presenter solves the differential equation d²y/dx² + 5dy/dx + 6y = e^(-2x). The equation is rewritten as (D² + 5D + 6)y = e^(-2x), where f(D) = D² + 5D + 6 and X = e^(-2x). The auxiliary equation is m² + 5m + 6 = 0, which factors to (m+2)(m+3) = 0, giving roots m = -2 and m = -3. Since the roots are real and distinct, the complementary function is C₁e^(-2x) + C₂e^(-3x).
Finding the Particular Integral When a Root Makes the Denominator Zero [32:30]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X = 1/(D² + 5D + 6) * e^(-2x). The presenter factors the denominator as (D+2)(D+3). Substituting D = -2 makes the (D+2) term zero, so instead, the presenter substitutes D = -2 into the (D+3) term, resulting in (-2+3) = 1. The particular integral then becomes e^(-2x)/(D+2). Using the formula for when the denominator is zero, the particular integral is x * e^(-2x)/1! = x * e^(-2x). The general solution is the sum of the complementary function and the particular integral: y = C₁e^(-2x) + C₂e^(-3x) + x * e^(-2x).
Example 5: d²y/dx² - 5dy/dx + 6y = 3e^(3x) and Modifying the Question [36:27]
The presenter solves the differential equation d²y/dx² - 5dy/dx + 6y = 3e^(3x). The presenter modifies the question to ensure a zero denominator case is demonstrated. The equation is rewritten as (D² - 5D + 6)y = 3e^(3x), where f(D) = D² - 5D + 6 and X = 3e^(3x). The auxiliary equation is m² - 5m + 6 = 0, which factors to (m-2)(m-3) = 0, giving roots m = 2 and m = 3. Since the roots are real and distinct, the complementary function is C₁e^(2x) + C₂e^(3x).
Finding the Particular Integral for the Modified Example 5 [39:20]
The presenter calculates the particular integral for the modified equation. The particular integral is 1/f(D) * X = 1/(D² - 5D + 6) * 3e^(3x). The presenter factors the denominator as (D-2)(D-3). Substituting D = 3 makes the (D-3) term zero, so instead, the presenter substitutes D = 3 into the (D-2) term, resulting in (3-2) = 1. The particular integral then becomes 3e^(3x)/(D-3). Using the formula for when the denominator is zero, the particular integral is x * 3e^(3x)/1! = 3x * e^(3x). The general solution is the sum of the complementary function and the particular integral: y = C₁e^(2x) + C₂e^(3x) + 3x * e^(3x).
Example 6: d²y/dx² + 4dy/dx + 4y = e^(-2x) and Real and Equal Roots [42:34]
The presenter solves the differential equation d²y/dx² + 4dy/dx + 4y = e^(-2x). The equation is rewritten as (D² + 4D + 4)y = e^(-2x), where f(D) = D² + 4D + 4 and X = e^(-2x). The auxiliary equation is m² + 4m + 4 = 0, which factors to (m+2)(m+2) = 0, giving roots m = -2 and m = -2. Since the roots are real and equal, the complementary function is (C₁ + C₂x)e^(-2x).
Finding the Particular Integral for Example 6 with Repeated Roots [45:16]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X = 1/(D² + 4D + 4) * e^(-2x). The presenter factors the denominator as (D+2)². Substituting D = -2 makes the (D+2)² term zero. Using the formula for when the denominator is zero and the power is 2, the particular integral is x²/2! * e^(-2x) = (x²/2) * e^(-2x). The general solution is the sum of the complementary function and the particular integral: y = (C₁ + C₂x)e^(-2x) + (x²/2) * e^(-2x).
Example 7: d²y/dx² + 4dy/dx + 4y = 5 + e^(-2x) and Multiple Terms [53:38]
The presenter solves the differential equation d²y/dx² + 4dy/dx + 4y = 5 + e^(-2x). The equation is rewritten as (D² + 4D + 4)y = 5 + e^(-2x), where f(D) = D² + 4D + 4 and X = 5 + e^(-2x). The auxiliary equation is m² + 4m + 4 = 0, which factors to (m+2)(m+2) = 0, giving roots m = -2 and m = -2. Since the roots are real and equal, the complementary function is (C₁ + C₂x)e^(-2x).
Finding the Particular Integral for Example 7 with Multiple Terms [55:19]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X = 1/(D² + 4D + 4) * (5 + e^(-2x)). The presenter splits the equation into two parts: 5/(D²+4D+4) + e^(-2x)/(D²+4D+4). For the first part, the presenter substitutes D = 0, resulting in 5/4. For the second part, the presenter uses the formula for when the denominator is zero, resulting in (x²/2) * e^(-2x). The general solution is the sum of the complementary function and the particular integral: y = (C₁ + C₂x)e^(-2x) + 5/4 + (x²/2) * e^(-2x).
Example 8: d²y/dx² + dy/dx - 6y = e^(3x) + e^(-3x) and Combining Terms [58:13]
The presenter solves the differential equation d²y/dx² + dy/dx - 6y = e^(3x) + e^(-3x). The equation is rewritten as (D² + D - 6)y = e^(3x) + e^(-3x), where f(D) = D² + D - 6 and X = e^(3x) + e^(-3x). The auxiliary equation is m² + m - 6 = 0, which factors to (m+3)(m-2) = 0, giving roots m = -3 and m = 2. Since the roots are real and distinct, the complementary function is C₁e^(-3x) + C₂e^(2x).
Finding the Particular Integral for Example 8 with Combined Terms [1:01:13]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X = 1/(D² + D - 6) * (e^(3x) + e^(-3x)). The presenter splits the equation into two parts: e^(3x)/((D+3)(D-2)) + e^(-3x)/((D+3)(D-2)). For the first part, the presenter substitutes D = 3, resulting in e^(3x)/(61) = e^(3x)/6. For the second part, the presenter substitutes D = -3, resulting in xe^(-3x)/(-5). The general solution is the sum of the complementary function and the particular integral: y = C₁e^(-3x) + C₂e^(2x) + e^(3x)/6 - (x/5) * e^(-3x).
Example 9: d²y/dx² + 4dy/dx + 5y = cosh(x) and Hyperbolic Functions [1:05:10]
The presenter solves the differential equation d²y/dx² + 4dy/dx + 5y = cosh(x). The equation is rewritten as (D² + 4D + 5)y = cosh(x), where f(D) = D² + 4D + 5 and X = cosh(x). The auxiliary equation is m² + 4m + 5 = 0. Since factorisation is not possible, the quadratic formula is used to find the roots: m = (-4 ± √(-4))/2 = -2 ± i. The roots are complex, in the form α ± iβ, where α = -2 and β = 1.
Finding the Particular Integral and General Solution for Example 9 [1:07:52]
The presenter finds the complementary function and particular integral for the equation. Since the roots are complex, the complementary function is e^(αx)(C₁cos(βx) + C₂sin(βx)) = e^(-2x)(C₁cos(x) + C₂sin(x)). The particular integral is 1/f(D) * X = 1/(D² + 4D + 5) * cosh(x). The presenter uses the formula cosh(x) = (e^(x) + e^(-x))/2 and splits the equation into two parts: e^(x)/(2(D²+4D+5)) + e^(-x)/(2(D²+4D+5)). Substituting D = 1 and D = -1, the particular integral becomes e^(x)/20 + e^(-x)/4. The general solution is the sum of the complementary function and the particular integral: y = e^(-2x)(C₁cos(x) + C₂sin(x)) + e^(x)/20 + e^(-x)/4.
Example 10: d²y/dx² - 6dy/dx + 9y = sinh(x) and Hyperbolic Sine [1:11:45]
The presenter solves the differential equation d²y/dx² - 6dy/dx + 9y = sinh(x). The equation is rewritten as (D² - 6D + 9)y = sinh(x), where f(D) = D² - 6D + 9 and X = sinh(x). The auxiliary equation is m² - 6m + 9 = 0, which factors to (m-3)(m-3) = 0, giving roots m = 3 and m = 3. Since the roots are real and equal, the complementary function is (C₁ + C₂x)e^(3x).
Finding the Particular Integral for Example 10 with Hyperbolic Sine [1:14:11]
The presenter calculates the particular integral for the equation. The particular integral is 1/f(D) * X = 1/(D² - 6D + 9) * sinh(x). The presenter uses the formula sinh(x) = (e^(x) - e^(-x))/2 and splits the equation into two parts: e^(x)/(2(D²-6D+9)) - e^(-x)/(2(D²-6D+9)). Substituting D = 1 and D = -1, the particular integral becomes e^(x)/8 - e^(-x)/32. The general solution is the sum of the complementary function and the particular integral: y = (C₁ + C₂x)e^(3x) + e^(x)/8 - e^(-x)/32.
Examples 11 and 12: Finding Particular Integrals Directly [1:16:22]
The presenter demonstrates finding particular integrals directly for two examples. For d²y/dx² - 9y = e^(3x), the particular integral is 1/(D²-9) * e^(3x). Factoring the denominator as (D-3)(D+3) and substituting D = 3, the particular integral becomes (x/6) * e^(3x). For d²y/dx² - 36y = e^(6x), the particular integral is 1/(D²-36) * e^(6x). Factoring the denominator as (D-6)(D+6) and substituting D = 6, the particular integral becomes (x/12) * e^(6x).
Practice Questions and Concluding Remarks [1:18:52]
The presenter provides several practice questions for viewers to solve, covering various cases discussed in the video. The presenter encourages viewers to practice these questions and refers them to the app for more materials. The presenter concludes by mentioning that the next parts of the series will cover the remaining methods to complete the unit.
