Question

Equations with the Independent Variable Missing. If a second order differential equation has the form \(y^{\prime \prime}=f(y, y)\), then the independent variable \(t\) does not appear explicitly, but only through the dependent variable \(y .\) If we let \(v=y^{\prime}\), then we obtain \(d v / d t=f(y, v)\). Since the right side of this equation depends on \(y\) and \(v\), rather than on \(t\) and \(t\), this equation is not of the form of the first order equations discussed in Chapter 2 . However, if we think of \(y\) as the independent variable, then by the chain rule \(d v / d t=(d v / d y)(d y / d t)=v(d v / d y)\). Hence the original differential equation can be written as \(v(d v / d y)=f(y, v) .\) Provided that this first order equation can be solved, we obtain \(v\) as a function of \(y .\) A relation between \(y\) and \(t\) results from solving \(d y / d t=v(y)\). Again, there are two arbitrary constants in the final result. In each of Problems 34 through 39 use this method to solve the given differential equation. $$ y^{\prime \prime}+y=0 $$

Short answer

Question: Express the relation between \(y\) and \(t\) for the given second-order differential equation: \(y^{\prime \prime} = -y\). Answer: The relation between \(y\) and \(t\) for the given differential equation is: \(\int\frac{1}{\sqrt{-2y^2 + 2C_1}}dy = t + C_2\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step-by-step solution

Introduce the new variable \(v\)

Define a new variable \(v\) such that \(v = y^{\prime}\). Our goal is to express \(v\) as a function of \(y\) and then solve the resulting first-order differential equation.

Use the chain rule and the given differential equation

Using the chain rule, we can write the second derivative of \(y\) with respect to \(t\) as: $$ \frac{d^2y}{dt^2}=\frac{d}{dt}\left(\frac{dy}{dt}\right)=\frac{d}{dt}(v)=\frac{dv}{dy}\cdot\frac{dy}{dt}=v\frac{dv}{dy}. $$ Substitute the given differential equation \(y^{\prime \prime}=-y\) into this expression: $$ v\frac{dv}{dy}=-y. $$

Solve the first-order differential equation

Now we have a first-order differential equation to solve: $$ v\frac{dv}{dy}=-y. $$ Separate variables and integrate both sides: $$ \int v\, dv = -\int y\, dy. $$ Integrating both sides, we get: $$ \frac{1}{2}v^2 = -\frac{1}{2}y^2 + C_1, $$ where \(C_1\) is an arbitrary constant.

Solve for \(v\)

We can now solve for \(v\) as a function of \(y\): $$ v=\pm\sqrt{-2y^2+2C_1}. $$

Solve for \(y\)

Now, we have the equation: $$ \frac{dy}{dt} = v(y) = \pm\sqrt{-2y^2+2C_1}. $$ We will only be considering the positive square root case for simplicity. Similarly, the negative square root case can be solved. Separate variables and integrate both sides: $$ \int\frac{1}{\sqrt{-2y^2 + 2C_1}}dy = \int dt. $$ Integrating both sides, we get: $$ \int\frac{1}{\sqrt{-2y^2 + 2C_1}}dy = t + C_2, $$ where \(C_2\) is another arbitrary constant.

Final solution

The final solution to the given differential equation is the relation between \(y\) and \(t\) given by: $$ \int\frac{1}{\sqrt{-2y^2 + 2C_1}}dy = t + C_2. $$ This solution contains two arbitrary constants, \(C_1\) and \(C_2\), as expected for a second-order differential equation.

Key concepts

Second Order Differential Equation

Second order differential equations are a fascinating class of mathematical equations. They involve the second derivative of a function, which represents how the rate of change of a quantity is itself changing. In essence, it gives us the acceleration in a context where the first derivative represents velocity. When dealing with physical systems, second-order equations are common, such as in the analysis of oscillations, like springs and pendulums.

The provided exercise introduces a form where the independent variable is missing, specifically: \( y'' + y = 0 \). This suggests that the behavior of the system depends only on the dependent variable \( y \) and its rate of change, rather than directly involving time \( t \). By making substitutions and rearrangements, we can transform this second-order differential equation into a form that is easier to solve, highlighting the beauty of differential equations manipulation.

First Order Differential Equation

To solve a second-order differential equation without the independent variable, we often transform it into a first-order differential equation. This can simplify the process considerably. In our problem, we introduced \( v = y' \), which describes the rate of change of \( y \).

By using the chain rule, we express the second derivative of \( y \) as a product involving \( v \) and its derivative: \( v \frac{dv}{dy} = -y \). This is now a first-order differential equation in terms of \( v \) and \( y \). First-order differential equations are usually more straightforward to solve because they only involve one derivative, making them a crucial step in breaking down complex equations.

Solving this first-order equation allows us to then reintroduce the time variable \( t \), to find the relationship between \( y \) and \( t \).

Variables and Integration

In the realm of differential equations, solving often involves separating variables and integrating. When we have our first-order equation \( v \frac{dv}{dy} = -y \), separating the variables is the first step. This means rearranging terms to isolate one variable on each side of the equation.

Here, we move terms around to get \( \int v\, dv = -\int y\, dy \). This separation allows us to tackle each side individually with integration, a fundamental mathematical operation used to determine accumulated values, like areas under curves or, in this case, solutions of equations.

The integrals give \( \frac{1}{2}v^2 = -\frac{1}{2}y^2 + C_1 \), introducing an arbitrary constant \( C_1 \), which accounts for the generality of antiderivatives. This step is key in finding more specific solutions to differential equations.

Arbitrary Constants

Arbitrary constants emerge naturally when integrating, representing the infinite number of solutions that exist when solving differential equations. In any integration step, an arbitrary constant \( C \) is added because differentiation of a constant is zero, which means the original function could have included any constant.

In our solution steps, we encountered \( C_1 \) and \( C_2 \). These constants remained undetermined initially because the problem is general. For a complete solution to a differential equation, boundary or initial conditions would provide specific values for these constants. They essentially tailor the general solution to fit particular scenarios.

In second-order differential equations, having two arbitrary constants makes sense since the solution needs to accommodate the initial values of both the function and its first derivative. This flexibility in solutions is what makes differential equations both powerful and broadly applicable across many fields.