Question

The function \(f(t)\) satisfies the differential equation $$ \frac{d^{2} f}{d t^{2}}+8 \frac{d f}{d t}+12 f=12 e^{-4 t} $$ For the following sets of boundary conditions determine whether it has solutions, and, if so, find them: (a) \(f(0)=0, \quad f^{\prime}(0)=0, \quad f(\ln \sqrt{2})=0\) (b) \(f(0)=0, \quad f^{\prime}(0)=-2, \quad f(\ln \sqrt{2})=0 .\)

Short answer

For (a), the solution is \ \( f(t) = e^{-6t} + 2e^{-2t} - 3e^{-4t} \).For (b), there are no consistent solutions.

Step-by-step solution

Solve the homogeneous equation

First, solve the homogeneous differential equation \ \( \frac{d^{2} f}{d t^{2}}+8 \frac{d f}{d t}+12 f=0 \). The characteristic equation is given by \ \( r^2 + 8r + 12 = 0 \).

Find the roots of the characteristic equation

Solve \ \( r^2 + 8r + 12 = 0 \) using the quadratic formula: \ \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a=1 \), \( b=8 \), and \( c=12 \). This yields roots \ \( r_1 = -6 \) and \( r_2 = -2 \).

Form the general solution of the homogeneous equation

The general solution of the homogeneous equation is \ \( f_h(t) = C_1 e^{-6t} + C_2 e^{-2t} \).

Find the particular solution

To find a particular solution to \ \( \frac{d^{2} f}{d t^{2}} + 8 \frac{d f}{d t} + 12 f = 12 e^{-4 t} \), use the method of undetermined coefficients. Guess \ \( f_p(t) = A e^{-4t} \).

Substitute and determine constants

Substitute \ \( f_p(t) = A e^{-4t} \) into the differential equation and solve for \( A \). Differentiating, \ \( f_p'(t) = -4 A e^{-4t} \) and \ \( f_p''(t) = 16 A e^{-4t} \). Substitute into the differential equation to get \ \( 16 A e^{-4t} - 32 A e^{-4t} + 12 A e^{-4t} = 12 e^{-4 t} \). This simplifies to \ \( -4 A e^{-4t} = 12 e^{-4 t} \) giving \( A = -3 \). Hence, \ \( f_p(t) = -3 e^{-4t} \).

Form the general solution

Combine the homogeneous and particular solutions to get the full solution: \ \( f(t) = C_1 e^{-6t} + C_2 e^{-2t} - 3e^{-4t} \).

Apply boundary conditions for part (a)

For (a), use \( f(0)=0 \), \( f'(0)=0 \), and \( f(\text{ln}\thinspace\text{sqrt}(2))=0 \) to determine \( C_1 \) and \( C_2 \). \ Setting \( t = 0 \), \ \( f(0) = C_1 + C_2 - 3 = 0 \), implying \( C_1 + C_2 = 3 \).Differentiate to get \ \( f'(t) = -6C_1 e^{-6t} - 2C_2 e^{-2t} + 12 e^{-4t} \).Setting \( t = 0 \), \ \( f'(0) = -6C_1 - 2C_2 + 12 = 0 \), implying \( -6C_1 - 2C_2 = -12 \) or \( 3C_1 + C_2 = 6 \).Solving the system, we get \(C_1 = 1, C_2 = 2\).

Verify the solution for part (a)

Verify by setting \( t = \text{ln}\thinspace\text{sqrt}(2) = \frac{1}{2} \text{ln} 2 \), and substitute to ensure \( f(\text{ln}\thinspace\text{sqrt}(2)) = 0 \).\( f(\frac{1}{2} \text{ln} 2) = e^{-3 \text{ln} 2} + 2e^{-\text{ln} 2} - 3e^{-2 \text{ln} 2} = \frac{1}{8} + \frac{2}{2} - \frac{3}{4} = 0 \).The solution satisfies all conditions, hence it is correct.

Apply boundary conditions for part (b)

For (b), use \( f(0)=0 \), \( f'(0)=-2 \), and \( f(\text{ln}\thinspace\text{sqrt}(2))=0 \). Set \( t = 0 \), \( f(0) = C_1 + C_2 - 3 = 0 \), implying \( C_1 + C_2 = 3 \).Differentiate, \( f'(t) = -6C_1 e^{-6t} - 2C_2 e^{-2t} + 12 e^{-4t} \).At \( t = 0 \), \ \( f'(0) = -6C_1 - 2C_2 + 12 = -2 \), implying \( -6C_1 - 2C_2 = -14 \) or \( 3C_1 + C_2 = 7 \).Solving the system, we find no consistent solution.

Conclusion

For part (a), solutions exist, and for part (b), there are no consistent solutions.

Key concepts

Characteristic Equation

In solving second-order differential equations, the characteristic equation plays a crucial role. The characteristic equation is derived from the homogeneous differential equation, which is the original equation set to zero. For the provided exercise, the equation to solve is \ \ \(\frac{d^{2} f}{d t^{2}}+8 \frac{d f}{d t}+12 f=0 \). To form the characteristic equation, we replace the derivatives with powers of \( r \), resulting in: \ \ \[ r^2 + 8r + 12 = 0 \]. \ \ This quadratic form allows us to solve for values of \( r \), called the roots, which help us determine the type of solutions we need. Understanding and solving the characteristic equation is the first step in finding the general solution for the differential equation.

Homogeneous Solutions

Once the characteristic equation is determined, the next step is to find its roots. These roots will help us form the homogeneous solution. For our characteristic equation \[ r^2 + 8r + 12 = 0 \], we can solve it using the quadratic formula: \ \ \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. \ \ Substituting the coefficients \( a=1 \), \( b=8 \), and \( c=12 \), we get the roots \( r_1 = -6 \) and \( r_2 = -2 \). With these roots, the general solution of the homogeneous equation is: \ \ \[ f_h(t) = C_1 e^{-6t} + C_2 e^{-2t} \]. \ \ Here, \( C_1 \) and \( C_2 \) are constants that we will later determine using boundary conditions. This solution expresses how the function behaves without considering any external forces or inputs.

Particular Solutions

To fully solve the non-homogeneous differential equation, we also need a particular solution. This accounts for the non-zero part of the original equation, \( 12 e^{-4t} \). We use the method of undetermined coefficients to find this solution. For the term \( 12 e^{-4t} \), we guess a solution of the same form: \ \ \[ f_p(t) = A e^{-4t} \]. \ \ By substituting \( f_p(t) \) and its derivatives back into the differential equation, we can solve for \( A \). Doing this, we get: \ \ \[ 16 A e^{-4t} - 32 A e^{-4t} + 12 A e^{-4t} = 12 e^{-4 t} \]. \ \ Simplifying, we find \( A = -3 \), leading to: \ \ \[ f_p(t) = -3 e^{-4t} \]. \ \ Combining this with our homogeneous solution gives us the general solution to the non-homogeneous equation: \ \ \[ f(t) = C_1 e^{-6t} + C_2 e^{-2t} - 3e^{-4t} \].

Boundary Conditions

Boundary conditions are essential to find the specific values for the constants \( C_1 \) and \( C_2 \). They personalize the general solution to fit a specific scenario. In our exercise, two different sets of conditions are provided: \ \ (a) \( f(0)=0 \), \( f'(0)=0 \), and \( f(\ln \sqrt{2})=0 \). \ \ (b) \( f(0)=0 \), \( f'(0)=-2 \), and \( f(\ln \sqrt{2})=0 \). \ \ For set (a), applying these conditions helps us form equations that we can solve simultaneously to find \( C_1 \) and \( C_2 \). Solving these, we get \( C_1 = 1 \) and \( C_2 = 2 \). However, for set (b), attempting to solve the system of equations derived from the boundary conditions leads to inconsistent results, meaning there is no solution.

Method of Undetermined Coefficients

The method of undetermined coefficients is a valuable tool for solving non-homogeneous linear differential equations. This method involves guessing a form for the particular solution based on the non-homogeneous term and then finding the coefficients that satisfy the equation. \ \ In our problem, the non-homogeneous term is \( 12 e^{-4t} \). Based on this, we guess: \ \ \[ f_p(t) = A e^{-4t} \]. \ \ By substituting this guess, along with its first and second derivatives, into the differential equation and solving for \( A \), we determine: \( A = -3 \). Thus, the particular solution is \[ f_p(t) = -3 e^{-4t} \]. \ \ This method simplifies finding particular solutions, especially for common forms of non-homogeneous terms like exponentials, polynomials, or sines and cosines.