Problem 40 - Differential Equations and Linear Algebra

Question

Determine the general solution to $$ y^{\prime \prime}-2 m y^{\prime}+\left(m^{2}-k^{2}\right) y=0 $$ where \(m\) and \(k\) are positive constants. Show that the solution can be written in the form: $$ y(x)=e^{m x}\left(c_{1} \cosh k x+c_{2} \sinh k x\right) $$

Step-by-Step Solution

Answer

The general solution of the given second-order linear homogeneous ODE is: $$ y(x) = c_1 e^{mx} \cosh kx + c_2 e^{mx} \sinh kx. $$

Step 1: Write down the given ODE and the characteristic polynomial

The given second-order ODE is: $$ y^{\prime\prime} - 2my^{\prime} + (m^2 - k^2)y = 0. $$ To solve this ODE, let's first find the characteristic polynomial: $$ r^2 - 2mr + (m^2 - k^2) = 0. $$

Step 2: Solve the characteristic polynomial

Now, let's solve the polynomial to find the roots. Given the quadratic equation, we can use the quadratic formula to solve for r: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. $$ Substituting the values of a, b, and c, we get: $$ r = \frac{2m \pm \sqrt{(2m)^2 - 4(m^2 - k^2)}}{2}. $$ Simplifying the terms, we obtain: $$ r = m \pm k. $$ So, there are 2 distinct real roots: \(r_1 = m + k\) and \(r_2 = m - k\).

Step 3: Write down the general solution concerning the roots

As we have 2 distinct real roots, the general solution of the given ODE can be written as: $$ y(x) = c_1 e^{(m+k)x} + c_2 e^{(m-k)x}. $$

Step 4: Rewrite the general solution in the desired form

Recall that the hyperbolic cosine and hyperbolic sine functions can be written as: $$ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2}. $$ Now, let's rewrite the general solution using the hyperbolic functions. We can analyze the terms one by one: $$ e^{(m+k)x} = e^{mx}\cdot e^{kx} $$ and $$ e^{(m-k)x} = e^{mx}\cdot e^{-kx}. $$ Now, let's express these terms in terms of the hyperbolic functions. Add them together and divide by 2: $$ \frac{e^{(m+k)x} + e^{(m-k)x}}{2} = e^{mx} \cosh kx. $$ In the same way, take their difference and divide by 2: $$ \frac{e^{(m+k)x} - e^{(m-k)x}}{2} = e^{mx} \sinh kx. $$ Now substitute these expressions back into the general solution: $$ y(x) = c_1 e^{mx} \cosh kx + c_2 e^{mx} \sinh kx. $$ The general solution of the given ODE has now been rewritten in the desired form.

Key Concepts

Characteristic Polynomial

The characteristic polynomial is a crucial part of solving linear homogeneous differential equations with constant coefficients. In the given exercise, we start with a second-order ordinary differential equation (ODE): This ODE can be translated into a characteristic polynomial, which is essentially a quadratic equation derived by assuming solutions of the form \( y = e^{rx} \), where \( r \) represents potential roots. Substituting this form into the differential equation, we get the polynomial: \[ r^2 - 2mr + (m^2 - k^2) = 0. \] By solving this polynomial, we find the values of \( r \) that lead to solutions of the differential equation. The characteristic polynomial effectively transforms the problem from one involving differential equations into one involving algebra, making it simpler to find the solutions of the ODE.

General Solution

The general solution of a differential equation represents the set of all possible solutions. Given the ability of the characteristic polynomial to yield roots \( r_1 \) and \( r_2 \), the general solution for a second-order ODE takes the form of a linear combination of exponential functions. In this problem, the general solution is expressed as: \[ y(x) = c_1 e^{(m+k)x} + c_2 e^{(m-k)x}, \] where \( c_1 \) and \( c_2 \) are arbitrary constants determined by initial or boundary conditions. This solution uses the roots derived from the characteristic polynomial. By manipulating these expressions further, we can convert them into an alternative form using hyperbolic functions, showcasing one of the flexible nature of these solutions.

Hyperbolic Functions

Hyperbolic functions \( \cosh x \) and \( \sinh x \) resemble the trigonometric functions cosine and sine but are defined using exponential functions. These are particularly useful in rewriting solutions to differential equations due to their unique properties. The hyperbolic cosine and sine are given by: - \( \cosh x = \frac{e^x + e^{-x}}{2} \) - \( \sinh x = \frac{e^x - e^{-x}}{2} \) In this exercise, we express the general solution in terms of hyperbolic functions: - Combining exponential terms gives \( e^{(m+k)x} + e^{(m-k)x} = 2e^{mx} \cosh kx \) - Similarly, \( e^{(m+k)x} - e^{(m-k)x} = 2e^{mx} \sinh kx \) Thus, the solution can be neatly rewritten to utilize these functions as: \[ y(x) = e^{mx}(c_1 \cosh kx + c_2 \sinh kx), \] highlighting the practical use of hyperbolic functions in simplifying differential equation solutions.

Quadratic Formula

The quadratic formula is a fundamental tool used to solve quadratic equations in the form \( ax^2 + bx + c = 0 \). It provides the roots of the equation by using the formula: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] In the context of solving the characteristic polynomial in the given exercise, we apply this formula to the polynomial \( r^2 - 2mr + (m^2 - k^2) = 0 \). Here, \( a = 1 \), \( b = -2m \), and \( c = m^2 - k^2 \). Substituting these into the quadratic formula results in: \[ r = \frac{2m \pm \sqrt{(2m)^2 - 4(m^2 - k^2)}}{2}, \] leading to simplification and the discovery of roots \( r_1 = m + k \) and \( r_2 = m - k \). Using the quadratic formula is critical as it allows us to easily find the roots of the characteristic polynomial, forming the basis for the general solution to the differential equation.