Problem 12 - Advanced Engineering Mathematics

Question

Solve the given differential equation by undetermined coefficients. \(y^{\prime \prime}-16 y=2 e^{4 x}\)

Step-by-Step Solution

Answer

The solution is \( y = c_1 e^{4x} + c_2 e^{-4x} + \frac{1}{4} x e^{4x} \).

Step 1: Write the homogeneous equation

First, consider the homogeneous equation associated with the given differential equation: \[ y'' - 16y = 0 \]

Step 2: Solve the homogeneous equation

The characteristic equation for the homogeneous equation is \[ r^2 - 16 = 0 \]Solving for \(r\), we get \[ r^2 = 16 \]which gives roots \(r = 4\) and \(r = -4\). Hence, the general solution for the homogeneous equation is \[ y_h = c_1 e^{4x} + c_2 e^{-4x} \] where \(c_1\) and \(c_2\) are constants.

Step 3: Formulate the particular solution assumption

Assume a particular solution of the form \[ y_p = A x e^{4x} \] because \(e^{4x}\) is a solution to the homogeneous equation and we already have \(e^{4x}\) terms in \(y_h\).

Step 4: Compute derivatives of the particular solution

Compute the first and second derivatives of \(y_p\):The first derivative is \[ y_p' = A e^{4x} + 4Ax e^{4x} \]The second derivative is \[ y_p'' = 8Ae^{4x} + 16Ax e^{4x} \]

Step 5: Substitute into the original differential equation

Substitute \(y_p\), \(y_p'\), and \(y_p''\) back into the left-hand side of the original differential equation:\[ y_p'' - 16y_p = (8A e^{4x} + 16Ax e^{4x}) - 16Ax e^{4x} \]\[ = 8A e^{4x} \]Setting this equal to the right-hand side of the original differential equation, \(2e^{4x}\), gives:\[ 8A e^{4x} = 2 e^{4x} \]

Step 6: Solve for coefficient A

Equate the coefficients: \[ 8A = 2 \]Solve for \(A\):\[ A = \frac{2}{8} = \frac{1}{4} \]

Step 7: Write the complete solution

The general solution of the differential equation is the sum of the homogeneous and particular solutions: \[ y = y_h + y_p = c_1 e^{4x} + c_2 e^{-4x} + \frac{1}{4} x e^{4x} \]

Key Concepts

Undetermined Coefficients

The method of undetermined coefficients is a straightforward technique used to find particular solutions to linear non-homogeneous differential equations. This method is best suited for equations where the non-homogeneous part (right-hand side) is a linear combination of functions like exponentials, sines, cosines, or polynomials.
To apply this method, you guess a form for the particular solution, referred to as the ansatz, and determine the unknown coefficients by substituting back into the original equation. The initial guess for the particular solution should be based on the type of function present on the right side of the differential equation:

• For an exponential function like \(e^{ax}\), try a solution of form \(Ae^{ax}\).
• For a polynomial \(x^n\), guess \(Ax^n + Bx^{n-1} + ...\).
• For a sinusoidal function like \(\sin(bx)\) or \(\cos(bx)\), consider \(A\sin(bx) + B\cos(bx)\).

In some situations, such as when the forms of the complementary solution already include your guess, you need to multiply by \(x\) to ensure the particular solution is distinct from the homogeneous solution.

Homogeneous Equation

A homogeneous equation is a differential equation defined as having zero on the right-hand side, meaning there is no external forcing function. An example of this is \(y'' - 16y = 0\). Solving this allows you to find the complementary (or homogeneous) solution, which is a part of the complete solution to the differential equation.
To solve the homogeneous equation, you first derive the characteristic equation. For a differential equation like \(y'' - 16y = 0\), replace the derivatives with powers of a variable \(r\) to get the characteristic equation: \(r^2 - 16 = 0\).
Solving the characteristic equation provides roots, which lead to the homogeneous solution. In this case, the roots are \(r = 4\) and \(r = -4\), resulting in the general solution \(y_h = c_1 e^{4x} + c_2 e^{-4x}\), where \(c_1\) and \(c_2\) are arbitrary constants.

Particular Solution

The particular solution of a non-homogeneous differential equation is a specific solution that satisfies the entire equation, including the non-zero right-hand side. Given an equation like \(y'' - 16y = 2e^{4x}\), the aim is to find a function \(y_p\) that satisfies the equation when substituted.
To find this, assume a form for \(y_p\) based on the type of function you see as the non-homogeneous part. Here, since the right-hand side is \(2e^{4x}\), and \(e^{4x}\) is part of the solution to the homogeneous equation, you assume \(y_p = Ax e^{4x}\). This adjustment ensures that \(y_p\) does not duplicate the terms in the homogeneous solution.
After guessing \(y_p\), compute its derivatives and substitute these back into the original equation. By matching coefficients with the non-homogeneous term, you solve for the unknowns—here, \(A\)—to find \(y_p = \frac{1}{4}xe^{4x}\).

Characteristic Equation

The characteristic equation is crucial in finding the homogeneous solution of differential equations. It is derived by assuming a solution of the type \(y = e^{rx}\) to a linear homogeneous differential equation. This approach transforms the differential equation into a simpler algebraic form.
For the example \(y'' - 16y = 0\), substituting \(y = e^{rx}\) translates into \(r^2 e^{rx} - 16 e^{rx} = 0\). On dividing by \(e^{rx}\) (since \(e^{rx} eq 0\)), you're left with \(r^2 - 16 = 0\).
Solving the characteristic equation \(r^2 - 16 = 0\) yields the roots \(r = 4\) and \(r = -4\). These roots determine the distinct exponential functions that form the general solution to the homogeneous equation: \(y_h = c_1 e^{4x} + c_2 e^{-4x}\), with \(c_1\) and \(c_2\) being constants yet to be defined by initial conditions.