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. $$ y y^{\prime \prime}+\left(y^{\prime}\right)^{2}=0 $$

Short answer

Question: Solve the given second-order differential equation and find the relationship between \(y\) and \(t\): \(yy^{\prime\prime} + (y^{\prime})^{2} = 0\). Answer: The relationship between \(y\) and \(t\) is given by \(y(\ln|y| - C) - y = t + k\), where \(C\) and \(k\) are constants of integration.

Step-by-step solution

Rewrite the equation in terms of \(v\) and \(y\)

Given: $$ yy^{\prime\prime} + \left(y^{\prime}\right)^{2} = 0 $$ Let \(v = y'\), so \(v' = y''\). Then, our equation becomes: $$ yv' + v^2 = 0 $$

Solve the first-order equation for \(v\)

We have: $$ y\frac{dv}{dy}=-v^2 \\ \frac{1}{v^2}\frac{dv}{dy}=-\frac{1}{y} $$ Now, we integrate both sides with respect to \(y\): $$ \int\frac{1}{v^2}\frac{dv}{dy}dy = -\int\frac{1}{y}dy \\ \int\frac{1}{v^2}dv = -\int\frac{1}{y}dy $$ The integrals give us: $$ -\frac{1}{v} = -\ln|y| + C \\ \frac{1}{v} = \ln|y| - C $$ Now, let's find \(v(y)\): $$ v(y) = \frac{1}{\ln|y| - C} $$

Find the relationship between \(y\) and \(t\)

We know that $$ \frac{dy}{dt} = v(y) = \frac{1}{\ln|y| - C} $$ Now, we separate the variables: $$ (\ln|y| - C)dy = dt $$ Integrate both sides with respect to their respective variables: $$ \int(\ln|y| - C)dy = \int dt $$ After integration, we obtain: $$ y(\ln|y| - C) - y = t + k $$ Thus, the relationship between \(y\) and \(t\) is given by: $$ y(\ln|y| - C) - y = t + k $$ where \(C\) and \(k\) are both constants of integration.

Key concepts

Independent Variable

The independent variable is a crucial concept in math and science. It represents the variable that operates freely, influencing other variables. In differential equations, it's the variable that changes independently. Typically denoted by symbols like \(t\), \(x\), or \(z\), it lets us observe how changes affect the dependent variable.

For example, in our exercise, the function is of \(y^{\prime\prime}=f(y,v)\) form. Here, \(t\) acts as the independent variable but is absent explicitly. Instead, changes are expressed through the dependent variable \(y\). Therefore, even though \(t\) doesn't explicitly appear, its influence is captured through \(y\) and its derivatives.

Understanding the role of the independent variable is vital. It allows us to simplify equations and transform the problem into more digestible parts, as seen by our tactic to express connective changes through \(y\).

• Reflects the primary variable of influence.
• Can be "hidden" in differential equations like our example.

Second Order Differential Equation

A second order differential equation involves the second derivative of a function. It is essential for modeling systems with changing rates, where acceleration or curvature comes into play, unlike first order that deals only with velocity or slope changes.

In the given exercise, the equation \(yy^{\prime\prime} + (y^{\prime})^{2} = 0\) is a second order differential equation. This is because it includes the term \(yy^{\prime\prime}\), which features the second derivative of \(y\). The goal here is to solve for \(y\) in terms of the independent variable by first simplifying the equation to a form without explicit \(t\).

Often, dealing with second order differential equations involves higher complexity. Substituting variables and reducing order step-by-step allows reaching a solution more manageable.

• Characterized by the presence of \(y^{\prime\prime}\)
• Captures dynamics involving acceleration or bends in data plots.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with compositions of functions. It allows us to differentiate the outer function relative to the inner function's derivative.

In this exercise, to transition from a differential in \(t\) to one in \(y\), the chain rule steps in. If we consider \(v = y'\), then realizing \(\frac{dv}{dt} = \left(\frac{dv}{dy}\right) \left(\frac{dy}{dt}\right) = v \frac{dv}{dy}\), we harness it to represent \(\frac{dv}{dt}\) using \(y\) variables exclusively, simplifying complexities.

Mastery of the chain rule can simplify challenging differentiation tasks. By untangling compound derivatives, it becomes easier to derive intricate expressions with ease.

• Breaks down complex differentiation processes.
• Supports conversion between variables, crucial in differential equations.

Integration

Integration is essential in solving differential equations, especially for finding solutions over continuous intervals. It represents the process of finding an integral or antiderivative, which is crucial for moving from derivative definitions to function specifics.

In the exercise, integration plays a key role in deriving solutions for \(v\) and subsequently \(y\). When separating and integrating, the exercise reaches \(\int \frac{1}{v^2} \, dv = -\int \frac{1}{y} \, dy\), leading to solutions like \(v(y) = \frac{1}{\ln|y| - C}\).

Ultimately, integration allows us to establish relationships between independent variables and their derivatives, resolving the differential equation into an understandable deterministic function.

• Transforms derivatives into function forms.
• Necessary for finding complete solutions hence resolving differential questions.