Question

Consider the simple differential equation \(y^{\prime \prime}=0\). (a) Obtain the general solution by successive antidifferentiation. (b) View the equation \(y^{\prime \prime}=0\) as a second order linear homogeneous equation with constant coefficients, where the characteristic equation has a repeated real root. Obtain the general solution using this viewpoint. Is it the same as the solution found in part (a)?

Short answer

Question: Using two different methods, find the general solution for the simple differential equation \(y^{\prime \prime} = 0\). Answer: The general solution for the given differential equation is \(y(x) = C_1 + C_2x\), where C_1 and C_2 are constants.

Step-by-step solution

(a) Successive Antidifferentiation

To find the general solution using successive antidifferentiation, we perform the following steps: 1. Integrate both sides of the equation with respect to x: \(\int y^{\prime \prime} dx = \int 0 dx\) This will give the first derivative of y: \(y' = C_1\) 2. Integrate both sides of the equation again: \(\int y' dx = \int C_1 dx\) This will give the general solution: \(y(x) = C_1x + C_2\) where C_1 and C_2 are constants.

(b) Second-order linear homogeneous equation (Characteristic Equation)

To solve the given equation \(y^{\prime \prime} = 0\) as a second-order linear homogeneous equation with constant coefficients, consider the equation in the following form: \(y^{\prime \prime} = 0y' + 0y\) From this form, we can write the characteristic equation: \(r^2 = 0\) Solving for r, we find a repeated real root: \(r = 0, 0\) For a second-order linear homogeneous equation with a repeated real root (r), the general solution can be written as: \(y(x) = (C_1 + C_2x)e^{rx}\) In this case, r = 0, and the general solution becomes: \(y(x) = (C_1 + C_2x)e^{0} = C_1 + C_2x\)

Comparison of Solutions

Comparing the general solutions obtained in parts (a) and (b): \(y(x) = C_1x + C_2\) (from part a) \(y(x) = C_1 + C_2x\) (from part b) The general solutions are the same when the constants are renamed. Thus, both methods give the same general solution for the given differential equation.

Key concepts

Antidifferentiation

Antidifferentiation is a fundamental concept in calculus that involves finding the original function from its derivative. Imagine you know the rate of change of a quantity and want to discover the actual quantity itself. This process is essentially the 'opposite' of differentiation.

In mathematical terms, if you have a function's derivative, say \( f'(x) \), antidifferentiation enables you to determine the original function \( f(x) \). To do this, you integrate \( f'(x) \). Importantly, when integrating, a constant of integration \( C \) is always included since there can be an infinite number of original functions differing by a constant.

Let's see how antidifferentiation works in practice with the simple differential equation \( y'' = 0 \):

• First, integrate \( y'' \) to find \( y' \): \[ y' = \int 0 \, dx = C_1 \]
• Next, integrate \( y' \) to get \( y \): \[ y = \int C_1 \, dx = C_1x + C_2 \]

Now, the equation \( y(x) = C_1x + C_2 \) represents the general solution for \( y'' = 0 \), with \( C_1 \) and \( C_2 \) being arbitrary constants.

Characteristic Equation

The characteristic equation is a crucial concept used to solve linear differential equations, especially second-order ones with constant coefficients. It transforms the differential equation into an algebraic equation, which can often be solved more straightforwardly.

Consider the second-order linear homogeneous differential equation \( y'' + ay' + by = 0 \). The assumed solution is generally in the form \( y = e^{rx} \), where \( r \) is a constant. Substituting this assumed solution into the equation, you generate a polynomial equation, known as the characteristic equation.

For our introductory example \( y'' = 0 \):

• Rewrite it as \( y'' + 0y' + 0y = 0 \).
• Assume \( y = e^{rx} \). Then \( y' = re^{rx} \) and \( y'' = r^2e^{rx} \).
• Substitute these into the differential equation, leading to \( r^2e^{rx} = 0 \).
• Solving \( r^2 = 0 \), gives a repeated real root \( r = 0, 0 \).

For repeated roots, the solution typically takes the form \( (C_1 + C_2x)e^{rx} \). In this case, \( r = 0 \), simplifying our solution to \( y(x) = C_1 + C_2x \), matching the solution from antidifferentiation.

Second-Order Linear Homogeneous Equations

Second-order linear homogeneous equations form a broad class of differential equations essential in modeling various physical systems, like mechanical vibrations or electrical circuits. They're called 'second-order' because they involve the second derivative, 'linear' because they can be prepared in a linear form, and 'homogeneous' since there's no standalone function added beside the derivatives.

Such an equation typically looks like \( ay'' + by' + cy = 0 \). Solving these involves detecting the nature of its roots, using the characteristic equation (which you learned about earlier). Depending on the roots, the general solution structure varies:

For the case considered here, \( y'' = 0 \) is quite straightforward yet illustrative:

• This turns into a characteristic equation \( r^2 = 0 \).
• Having repeated roots \( r = 0 \) leads to solutions of the form \( y(x) = (C_1 + C_2x)e^{rx} \), simplifying to \( y(x) = C_1 + C_2x \).

These equations are pivotal in understanding natural phenomena and engineering problems, making them essential in the study of differential equations. Whether through successive antidifferentiation or characteristic equations, their solutions help unravel complex behaviors in different applied contexts.