Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}-5 y^{\prime}+6 y=e^{2 x}\)

Short answer

Linear second-order non-homogeneous differential equation. Solution: \[ y = C_1 e^{2x} + C_2 e^{3x} + \frac{1}{3} x e^{2x} \].

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \[ y^{\backprime \backprime} - 5 y^{\backprime} + 6 y = e^{2x} \].This is a linear second-order differential equation with constant coefficients and a non-homogeneous term on the right-hand side.

Solve the Homogeneous Part

First, solve the homogeneous equation \[ y^{\backprime \backprime} - 5 y^{\backprime} + 6 y = 0 \].The characteristic equation for this is \[ r^2 - 5r + 6 = 0 \]. Solving this quadratic equation gives \[ r = 2 \] and \[ r = 3 \]. Therefore, the general solution to the homogeneous part is \[ y_h = C_1 e^{2x} + C_2 e^{3x} \].

Find a Particular Solution

To find a particular solution to the non-homogeneous equation \[ y^{\backprime \backprime} - 5 y^{\backprime} + 6 y = e^{2x} \], assume a solution of the form \[ y_p = Ax e^{2x} \].This choice is because \[ e^{2x} \] is already a solution to the homogeneous equation. Calculate the derivatives \[ y_p^{\backprime} = Ae^{2x}(2x + 1) \] and \[ y_p^{\backprime \backprime} = 4Ax e^{2x} + 2Ae^{2x} \]. Substitute \[ y_p, y_p^{\backprime}, y_p^{\backprime \backprime} \] back into the differential equation and solve for \[ A \].

Combine the Solutions

The general solution to the differential equation is \[ y = y_h + y_p \]. Therefore, \[ y = C_1 e^{2x} + C_2 e^{3x} + Ax e^{2x} \]. Substitute \[ Ax e^{2x} = e^{2x} \] from the previous step to find \[ A \].

Solve for the Coefficient A

By substitution, we get \[ 4Ax e^{2x} + 2Ae^{2x} - 5Axe^{2x} - 5Ae^{2x} + 6Ax e^{2x} = e^{2x} \]. Simplify to solve for \[ A \]. This gives \[ A = \frac{1}{3} \]. Hence, the particular solution is \[ y_p = \frac{1}{3} x e^{2x} \].

Finalize the General Solution

Combine the homogeneous and particular solutions to get the general solution:\[ y = C_1 e^{2x} + C_2 e^{3x} + \frac{1}{3} x e^{2x} \].

Key concepts

Differential Equations

A differential equation is an equation involving an unknown function and its derivatives. These equations are essential in modeling various physical and natural systems. They can describe anything from the motion of planets to population dynamics. Differential equations can be classified into several types:

• Ordinary Differential Equations (ODEs)
• Partial Differential Equations (PDEs)
• Linear and Non-linear Differential Equations
• Homogeneous and Non-homogeneous Differential Equations

This article focuses on a linear second-order differential equation which includes the second derivative of the unknown function.

Homogeneous Solutions

To solve a linear second-order differential equation, we often begin by solving the homogeneous equation. For the given problem, the homogeneous part is \[ y^{\backprime \backprime} - 5 y^{\backprime} + 6 y = 0 \]. This equation does not have any external forces acting on the system (i.e., the right-hand side is zero). The solution of this part gives us the complementary function or the homogeneous solution. To find this, we convert the differential equation into its characteristic form: \[ r^2 - 5r + 6 = 0 \]. Solving this quadratic equation yields the roots \[ r_1 = 2 \] and \[ r_2 = 3 \]. Thus, the homogeneous solution is a combination of these exponential terms, given by \[ y_h = C_1 e^{2x} + C_2 e^{3x} \], where \[ C_1 \] and \[ C_2 \] are constants determined by boundary or initial conditions.

Particular Solutions

After solving the homogeneous part, the next step is to find a particular solution to the non-homogeneous equation. In the given problem, the non-homogeneous term is \[ e^{2x} \]. To find a particular solution, we assume a form that resembles the non-homogeneous term. However, because \[ e^{2x} \] is already a solution to the homogeneous part, we assume the particular solution in the form \[ y_p = Ax e^{2x} \]. By substituting \[ y_p \] and its derivatives back into the differential equation, we solve for the coefficient \[ A \]. This process involves:

• Calculating the first and second derivatives of \[ y_p \]
• Substituting these into the original equation
• Matching coefficients to solve for \[ A \]

Finally, we find \[ A = \frac{1}{3} \], so the particular solution is \[ y_p = \frac{1}{3} x e^{2x} \].

Characteristic Equation

The characteristic equation is a crucial tool in solving linear differential equations with constant coefficients. It translates the differential equation into an algebraic equation, which is often much simpler to solve. For the given differential equation \[ y^{\backprime \backprime} - 5 y^{\backprime} + 6 y = 0 \], the characteristic equation is formed by assuming solutions of the form \[ y = e^{rx} \]. Substituting \[ y = e^{rx} \], \[ y' = re^{rx} \], and \[ y'' = r^2 e^{rx} \] into the homogeneous equation results in \[ r^2 - 5r + 6 = 0 \]. This characteristic equation can then be solved using algebraic methods such as factoring, completing the square, or the quadratic formula. The roots of the characteristic equation, denoted as \[ r_1 \] and \[ r_2 \], determine the form of the solution to the homogeneous equation. For the given problem, the roots were \[ r_1 = 2 \] and \[ r_2 = 3 \], leading to the homogeneous solution \[ y_h = C_1 e^{2x} + C_2 e^{3x} \].