Question

Show that the general solution of \(y^{\mathrm{iv}}-y=0\) can be written as $$ y=c_{1} \cos t+c_{2} \sin t+c_{3} \cosh t+c_{4} \sinh t $$ Determine the solution satisfying the initial conditions \(y(0)=0, y^{\prime}(0)=0, y^{\prime \prime}(0)=1\) \(y^{\prime \prime \prime}(0)=1 .\) Why is it convenient to use the solutions cosh \(t\) and sinh \(t\) rather than \(e^{t}\) and \(e^{-t} ?\)

Short answer

Using hyperbolic functions, such as \(cosh(t)\) and \(sinh(t)\), is more convenient when solving higher-order linear homogeneous differential equations with constant coefficients because they allow us to separate the components of solutions due to imaginary unit roots and real roots. This separation simplifies the process of solving for constants when applying initial conditions, making the overall solution process more straightforward.

Step-by-step solution

Find the complementary function

To solve the given differential equation \(y^{iv} - y = 0\), we need to find the complementary function. This involves finding the roots of the characteristic equation. Let's begin by forming this equation: $$ r^4 - 1 = 0 $$ Now, we can factor the equation as: $$ (r^2 + 1)(r^2 - 1) = 0 $$ Further factorizing, we get: $$ (r^2 + 1)(r + 1)(r - 1) = 0 $$ The roots are \(r_1 = 1\), \(r_2 = -1\), \(r_3 = i\), and \(r_4 = -i\). Given these roots, the complementary function is: $$ y_c(t) = c_1e^t + c_2e^{-t} + c_3e^{it} + c_4e^{-it} $$

Rewrite using Hyperbolic Trigonometric Functions

In order to make the problem easier to work with, we rewrite the complementary function using hyperbolic trigonometric functions. Recall that \(cosh(x) = \frac{e^x + e^{-x}}{2}\) and \(sinh(x) = \frac{e^x - e^{-x}}{2}\): $$ y_c(t) = c_1 cosh(t) + c_2 sinh(t) + c_3 cos(t) + c_4 sin(t) $$

Apply Initial Conditions

Now we apply the initial conditions \(y(0) = 0\), \(y'(0) = 0\), \(y''(0) = 1\), and \(y'''(0) = 1\). For \(y(0) = 0\): $$ 0 = c_1 cosh(0) + c_2 sinh(0) + c_3 cos(0) + c_4 sin(0) $$ This gives us \(c_1 + c_3 = 0 \implies c_3 = -c_1\). For \(y'(0) = 0\), we first find the \(y'(t)\): $$ y'(t) = c_1 sinh(t) + c_2 cosh(t) - c_3 sin(t) + c_4 cos(t) $$ Now applying the condition: $$ 0 = c_1 sinh(0) + c_2 cosh(0) - c_3 sin(0) + c_4 cos(0) $$ This gives us \(c_2 + c_4 = 0 \implies c_4 = -c_2\). For \(y''(0) = 1\), we first find the \(y''(t)\): $$ y''(t) = c_1 cosh(t) + c_2 sinh(t) - c_3 cos(t) - c_4 sin(t) $$ Now applying the condition: $$ 1 = c_1 cosh(0) + c_2 sinh(0) - c_3 cos(0) - c_4 sin(0) $$ This gives us \(c_1 - c_3 = 1 \implies c_1 = \frac{1}{2} \implies c_3 = -\frac{1}{2}\). For \(y'''(0) = 1\), we first find the \(y'''(t)\): $$ y'''(t) = c_1 sinh(t) + c_2 cosh(t) + c_3 sin(t) - c_4 cos(t) $$ Now applying the condition: $$ 1 = c_1 sinh(0) + c_2 cosh(0) + c_3 sin(0) - c_4 cos(0) $$ This gives us \(c_2 - c_4 = 1 \implies c_2 = \frac{1}{2} \implies c_4 = -\frac{1}{2}\).

Write the Solution

Now that we have found all the constants, we can write the solution: $$ y(t) = \frac{1}{2} cosh(t) + \frac{1}{2} sinh(t) - \frac{1}{2} cos(t) - \frac{1}{2} sin(t) $$

Conclusion

It is more convenient to use the hyperbolic functions \(cosh(t)\) and \(sinh(t)\) rather than \(e^t\) and \(e^{-t}\) because they allow us to consider separate components of solutions, one due to imaginary unit roots and the other due to real roots. This separation makes solving for constants using initial conditions much more straightforward.

Key concepts

Characteristic Equation

A crucial component in solving differential equations like \( y^{\mathrm{iv}} - y = 0 \) is finding the characteristic equation. This equation arises from assuming solutions of the differential equation have the form of \( e^{rt} \). By substituting and applying initial conditions, we convert the differential equation into an algebraic one: \[ r^4 - 1 = 0\] This characteristic equation is the starting point to determine the roots, which indicate the nature of the solution.

• Roots of characteristic equations can be real or complex.
• Complex roots lead to terms involving sine and cosine functions in the solution.
• Real roots correspond to exponential terms or hyperbolic functions.

In our case, the characteristic equation factors easily, revealing roots \( r_1 = 1 \), \( r_2 = -1 \), \( r_3 = i \), and \( r_4 = -i \). These roots are essential for constructing the complementary function, which is the next step for solving the differential equation.

Complementary Function

The complementary function is derived from the roots of the characteristic equation, and it represents the general solution of a homogeneous differential equation. For our equation \( y^{iv} - y = 0 \), after identifying the roots, the complementary function takes the form:\[y_c(t) = c_1 e^t + c_2 e^{-t} + c_3 e^{it} + c_4 e^{-it}\]This function incorporates solutions that correspond to each type of root:

• Real roots, such as \( r = 1 \) and \( r = -1 \), lead to exponential functions \( e^t \) and \( e^{-t} \).
• Complex roots \( r = i \) and \( r = -i \) give rise to trigonometric functions \( cos(t) \) and \( sin(t) \).

The combination of these functions forms the complementary function, which encompasses the different aspects of periodic and exponential behavior in the solution. However, to simplify calculations, we might prefer to express it in terms of hyperbolic functions as discussed in the next section.

Hyperbolic Functions

Hyperbolic functions, like \( \cosh(t) \) and \( \sinh(t) \), are especially useful in simplifying solutions to differential equations with complex roots. They are analogous to the trigonometric functions \( \cos(t) \) and \( \sin(t) \) but serve a unique purpose for hyperbolic geometries:

• \( \cosh(t) = \frac{e^t + e^{-t}}{2} \)
• \( \sinh(t) = \frac{e^t - e^{-t}}{2} \)

These representations are not just mathematical conveniences; they reflect the symmetry and growth properties inherent in certain differential equations with the use of real exponential terms.By rewriting the complementary function:\[y_c(t) = c_1 \cosh(t) + c_2 \sinh(t) + c_3 \cos(t) + c_4 \sin(t)\]This form provides a clearer picture while solving for unknown constants from initial conditions by distinctly separating behavior due to the real and imaginary parts of the roots.

Initial Conditions

Initial conditions are crucial in determining the specific solution to a differential equation from its general solution. They help pinpoint the values of arbitrary constants present in the complementary function. In our problem, the initial conditions given are:

• \( y(0) = 0 \)
• \( y'(0) = 0 \)
• \( y''(0) = 1 \)
• \( y'''(0) = 1 \)

Applying these conditions step-by-step involves differentiating the complementary function to obtain its derivatives and substituting \( t = 0 \) into each.1. **First**, substitute into the main function: \[ 0 = c_1 \, \cosh(0) + c_3 \, \cos(0) \] Helping us reduce the number of constants by setting relationships such as \( c_1 = -c_3 \). 2. **Then**, differentiate and substitute into the first derivative: \[ 0 = c_2 \, \cosh(0) + c_4 \, \cos(0) \] This informs us that \( c_2 = -c_4 \).3. **Next**, use the second derivative for non-zero condition values: \[ 1 = c_1 \, \cosh(0) - c_3 \, \cos(0) \] Solving this determines specific values like \( c_1 = \frac{1}{2} \).4. **Finally**, use the third derivative to refine these constants further: \[ 1 = c_2 \, \cosh(0) - c_4 \, \cos(0) \] Resulting in \( c_2 = \frac{1}{2} \).This iterative application ensures the specific solution satisfies the differential equation and adheres precisely to all initial conditions, ensuring a unique fit within the framework of the problem.