Find the general solution of each of the differential equations in exercise. (d^2 y)/(d x^2)+6 (

Find the general solution of each of the differential...

Key Concepts

Nonhomogeneous Differential Equation

A nonhomogeneous differential equation includes a forcing function or source term apart from terms involving the dependent variable and its derivatives. The overall solution to such an equation is the sum of the complementary (homogeneous) solution and one particular solution that accounts for the nonhomogeneous part.

Complementary Solution

Also known as the homogeneous solution, the complementary solution is derived by solving the associated homogeneous differential equation (i.e., setting the nonhomogeneous term to zero). Its structure is determined by the roots of the characteristic equation, and it represents the general solution to the homogeneous part of the differential equation.

Superposition Principle

The superposition principle states that the general solution of a linear nonhomogeneous differential equation can be written as the sum of the general solution of its homogeneous part and any particular solution to the nonhomogeneous equation. This principle exploits the linearity of the equation, allowing the separation of distinct contributions to the solution.

Linear Differential Equations with Constant Coefficients

These are differential equations where the dependent variable and its derivatives appear linearly, and the coefficients of these terms are constants. They are well-studied because their constant coefficients allow for the use of systematic methods, such as the characteristic equation, to derive general solutions.

Characteristic Equation

The characteristic equation is formed by replacing the derivatives in the homogeneous part of a linear differential equation with powers of a parameter (commonly r), resulting in an algebraic equation. The roots of this equation indicate the form of the homogeneous solution and reveal whether the solution consists of distinct real exponentials, repeated roots, or complex conjugates.

Method of Undetermined Coefficients

This method is used to find a particular solution for linear differential equations with constant coefficients when the nonhomogeneous term is of a specific form, such as exponentials, polynomials, or trigonometric functions. It involves guessing a form for the particular solution with unknown coefficients, substituting it into the differential equation, and solving for these coefficients.

Transcript

00:01 Dear students, here we have a differential equation of second order and we are asked to find the general solution.

00:10 So as it is a non -homogeneous equation, for our general solution, we need to find our complementary function as well as our particular integral, which is our particular solution.

00:27 So let's write our equation in the form of why, the number.

00:36 Prime plus 6 y prime plus 5y equals to 2 e to the power x plus 10 e to the power 5 x let's say this is our first equation now we can write the auxiliary equation as f of m equals to 0 so our auxiliary equation will be m square plus 6m plus 5 equals to 0 so you can solve this by this equation by vectorization by middle term breakdown as m square plus 5m plus m plus 5 equals to 0 so by taking m common we get m plus 5 and by taking one common we'll get m by taking one common we'll get m plus 5 equals to 0 so our factors will be m plus 1 into m plus 5 equals to 0 so roots are m equals minus 1 and minus 5s so we've got distinct and real roots so we can write our complementary as yc equals to c1 e to the power minus x plus c2 e to the power minus 5 x so this is our y c now we want to find y p y p so now y p so y p can be written as you need to find y p so y p can be 2 e to the power x plus 10 e to the power 5x on our right -hand side so write it as a any let's say any constant term a e to the power x plus b e to the power 5x so this is according to the formula of writing er y p so now by taking the first derivative for y p will get a e to the power x plus of 5 b e to the power 5 x now by taking the second derivative we'll get the derivation of our e to the power exponential function is the same and the same with 25 exponential function will give us the same and by taking the derivation of 5x we will get 5 so if we multiply 5 by 5 we'll get 25 in to p e to the power 5x so this we have found our y prime and y double prime so we can put these values in our first equation putting the values of y prime y p and y prime of p and y double prime of p in equation one so we get we will get as a wide double prime is a e to the power x plus 25 b e to the power 5 x and plus 6 into y prime we'll get plus 6 a x and by multiplying 6 with 5 we'll get plus of 30 b e to the power 5x and plus of 5 y so, multiplying by 5 with our y, we get plus of 5 a x plus 5 b e 5 x which is equal to our right -hand side which is 2 e to the power x plus 10 e to the power 5x.

06:04 So now we can simplify this right by adding the terms of e to the power a e to the power x will get 12.

06:14 A e to the power x and the same for b e to the power x we'll get by adding the terms of b e to the power x will get 60 as we have got 30 and 25 and plus of 5 b e to the power x will get 60 b e to the power 5 x which will be equal to 2 e to power x plus 10 e to the power 5 so now by comparing the coefficients of, well first compare the coefficients of x, e to the power x...

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