Chapter 8: Problem 9
Obtain the general solution. $$ \left(D^{2}-4\right) y=e^{2 x}+2 $$
Short Answer
Expert verified
The general solution is \(y = C_1e^{2x} + C_2e^{-2x} + \frac{1}{4}xe^{2x} + \frac{2}{4}\).
Step by step solution
01
Identify the Differential Operator Equation
The given equation is \((D^2 - 4)y = e^{2x} + 2\), where \(D^2\) represents the second derivative of \(y\) with respect to \(x\). The expression \(D^2 - 4\) is the differential operator acting on \(y\).
02
Construct the Auxiliary Equation
The corresponding homogeneous equation is \((D^2 - 4)y = 0\). To solve the homogeneous part, we form the characteristic equation by replacing \(D\) with \(r\), resulting in \(r^2 - 4 = 0\).
03
Solve the Characteristic Equation
Solve \(r^2 - 4 = 0\) for \(r\). Factoring gives \((r-2)(r+2) = 0\), which gives roots \(r = 2\) and \(r = -2\).
04
Write the Complementary Solution
The complementary solution \(y_c\) for the homogeneous equation is found to be \(y_c = C_1e^{2x} + C_2e^{-2x}\), where \(C_1\) and \(C_2\) are arbitrary constants.
05
Find a Particular Solution for the Inhomogeneous Equation
We need a particular solution \(y_p\) for the non-homogeneous equation \((D^2 - 4)y = e^{2x} + 2\). Since \(e^{2x}\) is a solution of the homogeneous equation, we try \(y_p = Axe^{2x} + B\).
06
Determine Coefficients for Particular Solution
Substitute \(y_p = Axe^{2x} + B\) into \((D^2 - 4)y = e^{2x} + 2\) and find the derivatives: \(y_p' = Ae^{2x} + 2Axe^{2x}\) and \(y_p'' = 4Ae^{2x} + 4Axe^{2x}\). Substituting into the equation gives: \((4Ae^{2x} + 4Axe^{2x} - 4Axe^{2x}) = e^{2x} + 2\). Simplify to get \(4Ae^{2x} = e^{2x} + 2\).
07
Solve for the Coefficient A
Equate coefficients to solve for \(A\). Thus, \(4A = 1\) and \(B-4B=2\) which implies \(B=-1/2\). Therefore, \(A = \frac{1}{4}\).
08
Write the General Solution
Combine the complementary and particular solutions to write the general solution: \(y = C_1e^{2x} + C_2e^{-2x} + \frac{1}{4}xe^{2x} + \frac{2}{4}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Second Order Differential Equations
Second order differential equations involve the second derivative of a function. In simpler terms, this means that the rate of change of the rate of change (or curvature) of a function is being described. These equations can be written in a general form like \(aD^2y + bDy + cy = f(x)\), where \(D^2y\) is the second derivative, \(Dy\) is the first derivative, \(y\) is the function, and \(f(x)\) is some function of \(x\). If the right side, \(f(x)\), is zero, it becomes a homogeneous equation. Otherwise, it is non-homogeneous.
The main challenge when solving second order differential equations is to find a solution \(y(x)\) that satisfies the equation for all values of \(x\). These types of equations frequently model physical phenomena where acceleration, or change in speed, is a factor like mechanical vibrations and electrical circuits.
The main challenge when solving second order differential equations is to find a solution \(y(x)\) that satisfies the equation for all values of \(x\). These types of equations frequently model physical phenomena where acceleration, or change in speed, is a factor like mechanical vibrations and electrical circuits.
Characteristic Equation
When faced with a second order differential equation, one common approach is to solve for the characteristic equation. This is done by replacing the differential operator \(D\) in the equation with a variable, often \(r\), to form a polynomial equation.
- The given form \(D^2 - 4\) becomes \(r^2 - 4\).
- This simplification makes it easier to solve for \(r\).
- The roots \(r_1\) and \(r_2\) found from this polynomial will shape the general nature of the solution.
Homogeneous Equations
Homogeneous equations are a type of differential equation where the right-hand side is zero, meaning \(f(x) = 0\). Such equations only depend on the function \(y\) and its derivatives. The general form looks like \(aD^2y + bDy + cy = 0\).
To find a solution to the homogeneous equation, we use the characteristic equation and its roots. Since the characteristic equation \(r^2-4=0\) yields the roots \(r = 2\) and \(r = -2\), the solution to our homogeneous equation can be written as:
To find a solution to the homogeneous equation, we use the characteristic equation and its roots. Since the characteristic equation \(r^2-4=0\) yields the roots \(r = 2\) and \(r = -2\), the solution to our homogeneous equation can be written as:
- \(y_c = C_1e^{2x} + C_2e^{-2x}\).
- \(C_1\) and \(C_2\) are constants determined by initial conditions.
Particular Solution
While the homogeneous solution accounts for some parts of the system, finding a particular solution is necessary to incorporate the effects of external forces or inputs represented by \(f(x)\). The final solution to a differential equation is a combination of the homogeneous solution and a particular solution tailored to the non-homogeneous part.
For the equation \((D^2 - 4)y = e^{2x} + 2\), a particular solution was proposed as \(y_p = Axe^{2x} + B\), since \(e^{2x}\) appears in the non-homogeneous term and aligns with an existing homogeneous solution.
For the equation \((D^2 - 4)y = e^{2x} + 2\), a particular solution was proposed as \(y_p = Axe^{2x} + B\), since \(e^{2x}\) appears in the non-homogeneous term and aligns with an existing homogeneous solution.
- Derivative calculations and substitutions lead to determining parameters \(A\) and \(B\).
- Setting up equations from coefficients gives \(A=\frac{1}{4}\) and \(B=-\frac{1}{2}\).