Linear Higher Order Differential Equation | CF & PI |Lecture-I

The video introduces first-order differential equations, focusing on various methods for solving them. Key techniques discussed include variable separation, homogeneous equations, linear reducible to linear forms, and exact solutions. The presenter transitions into a new topic: higher-order differential equations with constant coefficients.

Linear higher-order differential equations with constant coefficients can have orders as high as ten, but the key feature is that their coefficients remain constant. The symbolic form uses capital D to represent derivatives; for example, dy/dx becomes D and d²y/dx² becomes D². To solve these equations, one must find the complementary function (C.F.) and particular integral (P.I.), where C.F. is derived from an auxiliary or characteristic equation by substituting d with m. If Q equals zero in any problem, only C.F will be present while P.I will be absent.

To find the complementary function (C.F) of a differential equation, first replace d with m to derive the auxiliary equation. For real and distinct roots, such as in the example where m^2 - 3m + 2 = 0 yields roots m=1 and m=2, C.F is expressed as c1 e^x + c2 e^(2x). When dealing with repeated real roots like in d^2 - 4d + 4 = 0 resulting in double root at m=2, C.F takes the form c1 + c2 x e^(2x). In cases of complex or imaginary roots from an equation like d^2 - d + 4 = 0 leading to half ± √15i/₂ for its solutions, C.F incorporates both exponential functions for real parts and trigonometric functions for imaginary components.

In this segment, the focus is on replacing 'd' with 'm' in a set of questions. The auxiliary equation derived from this process indicates that the values for 'm' are 1 and 2. Consequently, the characteristic function (C.F) resulting from these values can be expressed as c1 e^x + c2 e^(2x).

The problem involves applying a mathematical concept where 'm' is factored out, resulting in the value of m being ±2i. This leads to both real and imaginary roots. The complementary function (C.F) derived from these roots is expressed as y = e^(2x) + C_0 + C_2 cos(2x) + C_3 sin(2x). Given that the highest order of the equation is 3, three constants are introduced: c1, c2, and c3.

The equation presented is of the fourth order, leading to four constants. By factoring out m squared, the values for m are determined as 0 (repeated), 1, and -3. The repeated roots indicate that the complementary function includes these values accordingly.

The discussion revolves around solving differential equations with varying roots. The auxiliary equation yields values of m as 1 and ±i, indicating both real and imaginary roots are repeating. The complementary function (C.F.) is derived from these roots, resulting in a specific formula involving exponential and trigonometric functions. Additionally, the concept of particular integrals (P.I.) is introduced where different functions like e^x or sin/cos x replace constants to derive solutions.

To find the Particular Integral (P.I.) of a differential equation in symbolic form, start by expressing it as y = q/f(d), where f(d) is placed in the denominator. If q equals e^(ax), replace f(d) with f(a), ensuring that f(a) does not equal zero. If it does equal zero, multiply the numerator by x and differentiate the denominator until you obtain a non-zero value for its derivative at 'a'. This method continues iteratively to derive P.I., which has been demonstrated through solved examples.

To solve the equation involving e^(4x), we first identify the complementary function (C.F.) and then find the particular integral (P.I.). The formula for P.I. is y = q, where specific values are substituted based on previous calculations. By applying a rule that correlates coefficients with derivatives, we replace 'd' with 4 to derive our final answer.

The problem involves calculating the complementary function (C.F) and particular integral (P.I) for an equation involving e raised to a power. By substituting d with -2x, all values yield -8; however, when using positive 2x instead, results alternate between 8 and -8. A critical point arises when the denominator equals zero; in such cases, differentiation concerning d is necessary to resolve it. The final expression for P.I becomes x multiplied by e raised to 2x divided by 8. Ultimately, if one side of the equation equates to zero only then can C.F be determined alongside both components being calculated based on given exponential values.

When the right-hand side of a differential equation involves sine or cosine functions, specific rules apply for finding the particular integral. The solution can incorporate both even and odd powers, represented symbolically with derivatives like d² and d. For instance, when using sine in an example, replacing d² with negative a² leads to a valid solution approach. This method remains consistent even if cosine is used instead of sine.

To solve the equation, first determine the complementary function (C.F) using the auxiliary equation with roots 3 and -1. The particular integral (P.I) is derived by substituting cos(2x), replacing d² with negative square during simplification. Rationalization is necessary for clarity in calculations, leading to a final solution that combines both C.F and P.I.

In this example, the variable 'd' is replaced with 'm', yielding values of m as 1 and 2. The complementary function (C.F) is established, followed by finding the particular integral (P.I). To simplify calculations involving sine and cosine functions, a common factor of negative is applied. By multiplying both numerator and denominator by (3D - 7), the equation simplifies to yield y = differentiation of sin(3x), resulting in an expression combining cos(3x) and sin(3x). Ultimately, adding C.F to P.I provides the final answer for two questions based on cosine and sine rules.

To find the Particular Integral (P.I.) for functions involving cos, apply the cosine rule. When dealing with polynomial expressions like x² + x or higher powers, identify and extract the least degree term as common. Utilize both binomial expansion and these concepts to solve specific problems effectively. Two example questions will illustrate this process clearly.

To solve the equation, start by writing the auxiliary equation and substituting values for m, specifically -1 and 3. The complementary function (C.F) and particular integral (P.I) can be derived from this process. If confusion arises during exams regarding formulas, remember to alternate signs when dealing with positive and negative terms. Since x has a maximum degree of two in this context, only differentiate x squared twice while neglecting higher-order terms.

To find the P.I. of a combined function, start by factoring out 'm', which yields values of zero and negative one. The process begins with substituting d with negative one in the exponential term, leading to a denominator that becomes zero; thus, multiply the numerator by x and differentiate it repeatedly until resolving terms involving cosine. For quadratic terms like x squared, factor out 'd' while applying concepts such as 1 plus x inverse for simplification. Finally, integration is performed on all resulting expressions to derive the complete answer.

The final answer for solving Linear Differential Equations with Constant Coefficients is the sum of Complementary Function (C.F) and Particular Integral (P.I). The C.F can be derived from the auxiliary equation's roots: if they are real and distinct, it follows a specific exponential form; if equal, it includes an additional linear term. For imaginary roots, a different approach applies. Additionally, methods to compute P.I were discussed using examples like integrating e^x. Feedback on these lessons is encouraged to reach more students.