Question

Find the general solution of each differential equation. When an initial condition is given, find the particular solution satisfying that condition. a. \(y^{\prime \prime}-3 y^{\prime}+2 y=20 e^{-2 x}, \quad y(0)=0, \quad y^{\prime}(0)=6\). b. \(y^{\prime \prime}+y=2 \sin 3 x\). c. \(y^{\prime \prime}+y=1+2 \cos x\). d. \(x^{2} y^{\prime \prime}-2 x y^{\prime}+2 y=3 x^{2}-x, \quad x>0\).

Short answer

a) \(y = e^x - 10e^{2x} + 16 x e^{2x}\), b) \(y = -\frac{2}{5} \cos(3x) + \frac{\sin(3x)}{10} + C_1 \cos(x) + C_2 \sin(x)\), c) \(y = -x\sin(x) + \sin(x) + \cos(x) + C_1 \cos(x) + C_2 \sin(x)\), d) \(y = x(3x - 1)\)

Step-by-step solution

Solution for part a, step 1: Complementary function

First, let's find the complementary function of the given differential equation. To do this, we solve the homogeneous equation \(y''-3y'+2y=0\). This has the solutions \(y_h = Ae^{2x} + Be^{x}\) where A and B are constants.

Solution for part a, step 2: Particular integral

To find a particular solution of the original inhomogeneous equation, we guess a solution of the form \(y_p=Ce^{-2x}\) (because of the \(e^{-2x}\) term on the right-hand side). Substituting this guess into the inhomogeneous equation, we can solve for C to get a particular solution.

Solution for part a, step 3: Apply initial conditions

We combine the complementary function and the particular integral to form the general solution, then we apply the initial conditions to this general solution to find the coefficients A and B, and thus the particular solutions.

Solution for part b and c, step 1: Homogeneous solutions

The solutions to the homogeneous equations \(y'' + y = 0\) are \(y_h = A\cos(x) + B\sin(x)\).

Solution for part b and c, step 2: Particular integral

For the inhomogeneous equations, we guess solutions of the form \(y_p = C\cos(3x) + D\sin(3x)\) for part b and \(y_p = Cx + D\cos(x) + E\sin(x)\) for part c and substitute this guess into the inhomogeneous equation to find the constants.

Solution for part d, step 1: Homogeneous solution

The homogeneous equation is \(x^2y'' - 2xy' + 2y = 0\), a form of the Euler differential equation. The solution to this equation is of the form \(y = x^n\), where \(n\) is a solution of the characteristic equation. We then use this result to find the solution of the inhomogeneous Euler differential equation.

Key concepts

Complementary Function Differential Equation

Understanding the complementary function is integral to solving non-homogeneous differential equations. Consider a differential equation that can be split into a homogeneous part and an inhomogeneous part. The complementary function, denoted as y_h, represents the solution to the homogeneous equation. It encapsulates all possible solutions generated by the associated homogeneous equation without the external forcing term (like the right-hand side in an inhomogeneous equation).

To determine the complementary function, we solve the characteristic equation, typically a polynomial whose roots indicate the form of y_h. For example, for the equation y'' - 3y' + 2y = 0, the characteristic equation is r^2 - 3r + 2 = 0, which has roots that lead to the complementary function of the form Ae^{2x} + Be^{x}, where A and B are arbitrary constants that will be later determined by initial conditions.

Particular Integral Solution

The particular integral solution, often denoted as y_p, is the part of the general solution to a non-homogeneous differential equation that accounts for the structure of the non-homogeneous term. If the differential equation includes an inhomogeneous term, like 20e^{-2x}, we need to guess a form for y_p that resembles this inhomogeneous part and substitute that into the equation to find the specific parameters.

The method of undetermined coefficients or the method of variation of parameters are common techniques used to guess the form of y_p. For instance, with an inhomogeneous term like e^{-2x}, we might guess a solution of the form Ce^{-2x} and solve for C. This solution would not include the arbitrary constants found in the complementary function as it specifically addresses the inhomogeneous part of the differential equation.

Initial Conditions in Differential Equations

Initial conditions are crucial for determining the particular solution to a differential equation. They provide specific values of the function or its derivatives at a certain point. Given an initial condition such as y(0) = 0 and y'(0) = 6, we use these to solve for the arbitrary constants in the general solution, which is the sum of the complementary function y_h and the particular integral y_p.

For our example, after obtaining the general solution, the initial conditions allow us to plug in x = 0 and equate the resulting equation to the provided initial values, resulting in a system of equations to solve for A and B. This process ensures that our final solution not only satisfies the differential equation but also adheres to the given constraints, making it the unique applicable solution.

Euler Differential Equation

Euler differential equations, or sometimes called Cauchy-Euler equations, have variable coefficients and are of special form, typically appearing as x^2y'' - axy' + by = 0. These equations are powerful because they have solutions that can be solved using a characteristic equation method. The solution to an Euler equation is of the form x^n, where n is a value we solve for by substituting x^n into the Euler equation and deriving a polynomial called the characteristic equation.

For instance, in the equation x^2y'' - 2xy' + 2y = 0, substituting y = x^n would lead us to the characteristic equation, which once solved, tells us the powers of x that make up the solution. The general solution for the Euler equation is a combination of these findings, allowing us to address problems with variable coefficients smoothly.

Characteristic Equation Solution

The characteristic equation solution method is one of the foundational approaches to solving linear homogeneous differential equations with constant coefficients. After reducing a differential equation to its complementary form, we create an auxiliary or characteristic polynomial whose roots indicate the nature and form of the solution.

For example, with an equation like y'' + 3y' + 2y = 0, the characteristic equation is r^2 + 3r + 2 = 0. Solving this yields roots that determine whether solutions are real and distinct, real and repeated, or complex. Each type of root leads to specific forms of the homogeneous solution: exponential functions for real roots, and combinations of sines and cosines for complex roots. This characteristic solution is a crucial aspect when constructing the complementary function, ultimately forming half of the general solution to our differential equation.