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. $$ y y^{\prime \prime}-\left(y^{\prime}\right)^{3}=0 $$

Short answer

Based on the solution above, summarize the steps taken to solve the given second-order differential equation and provide the final equation relating y(t) to the original equation. 1. Introduced a new variable \(v=y'\) and expressed the second-order differential equation with this new variable. 2. Used the chain rule to change the independent variable to \(y\). 3. Formed a first-order equation with v and y. 4. Solved the first-order equation to obtain v as a function of y. 5. Obtained a relation between y and t by solving another first order equation. The final equation relating \(y(t)\) to the original equation is: $$ y(t) = \int -\frac{1}{\ln{y} + C_1} dt $$

Step-by-step solution

Introduce a New Variable #

Let \(v = y'\), then \(y'' = \frac{d v}{d t}\), and we want to express the given second-order differential equation with this new variable. $$ y \frac{d v}{d t} - (v)^3 = 0 $$

Use the Chain Rule #

Now, use the chain rule to change the independent variable to \(y\). By applying the chain rule, we obtain $$ y \cdot v \frac{d v}{d y} - v^3 = 0 $$

Form a First Order Equation #

Rearrange the equation into the form \(v \frac{d v}{d y} = f(y,v)\), in our case, this would be $$ v \frac{d v}{d y} = v^3 / y $$

Solve the First Order Equation #

Divide both sides by \(v^2\) (assuming \(v\neq0\)): $$ \frac{d v}{d y} = \frac{v^2}{y} $$ Now, integrate with respect to \(y\): $$ \int \frac{1}{v^2} dv = \int \frac{1}{y} dy $$ This results in: $$ -\frac{1}{v} = \ln{y} + C_1 $$

Obtain v as a Function of y #

Isolate \(v\), then we have: $$ v = y' = -\frac{1}{(\ln{y} + C_1)} $$

Obtain a Relation between y and t #

Now we want to find a relation between \(y\) and \(t\), so we solve the following first order equation: $$ \frac{dy}{dt} = -\frac{1}{(\ln{y} + C_1)} $$ Integrate both sides with respect to \(t\): $$ \int dy = -\int\frac{1}{\ln{y} + C_1} dt $$ This yields: $$ y(t) = \int -\frac{1}{\ln{y} + C_1} dt $$ Unfortunately, the integral on the right-hand side is not elementary; however, at this point, we have reduced the second-order differential equation to a first-order equation involving \(y(t)\). In general, this is the best that can be done analytically, and the relation between \(y\) and \(t\) may be found numerically if needed.

Key concepts

Chain Rule

Understanding the chain rule is vital when dealing with differential equations, particularly when an equation involves composite functions. In the context of differential equations, the chain rule enables us to change the variable of differentiation when one variable is a function of another. It's analogous to finding the gears within a clock that interlink; one turn of a small gear effects the turn of a larger one. In mathematical terms, if you have a function that is a composition of two functions, say \( y(t) = (f \( g(t) \) ) \), the rate at which \( y \) changes with respect to \( t \) depends on the rate at which \( g \) changes with respect to \( t \) and how \( f \) changes with respect to \( g \).

Formally, if \( v = \frac{dy}{dt} \), and you want to find \( \frac{dv}{dt} \), you can use the chain rule to express this as \( v \frac{dv}{dy} \). This idea was crucial in the problem statement, enabling us to express our second-order differential equation in a way that opens the door to integrating and solving it. The chain rule is not just a tool but rather a bridge that connects the rates of change between different layers of functions.

First Order Differential Equation

A first order differential equation is one of the most basic types of equations in the study of differential equations. It involves the derivatives of a function up to the first order only. The general form is often written as \( \frac{dy}{dx} = f(x, y) \), with \( y \) being the unknown function we're trying to solve for, and \( x \) usually represents an independent variable.

When simplifying a higher-order differential equation to a first-order one, as we've done in the step-by-step solution, we are reducing the complexity of the problem. Once we reduce the second-order equation to a first order, we can then leverage various methods - such as variable separation or integration techniques - to find solutions. First-order equations often unveil a direct relationship between variables and their rate of change, making them less complex and more straightforward to solve or approximate.

Integration Techniques

Solving differential equations often requires integration – the process of finding the original function given its derivative. It is essentially the reverse operation of differentiation and is a crucial part of calculus. Integration techniques can range from basic power rule integrals to more advanced methods such as integration by parts, trigonometric integrals, and partial fractions.

In our problem, we used a straightforward integration approach once we isolated \( \frac{dv}{dy} \). By integrating both sides of the resulting first-order equation, we were able to find an implicit solution for \( v \), our intermediate variable. Advanced integration techniques are needed when dealing with more complex functions, and a good grasp of these methods can be the difference between a problem being solvable or not. Learning when and how to apply different integration strategies is a powerful skill in solving various types of differential equations.

Variable Separation

Variable separation is an elegant method employed to solve certain types of first-order differential equations. The core idea is to manipulate the equation in such a way that all terms involving the dependent variable, \( y \), are on one side of the equation, and all terms involving the independent variable, \( x \), are on the other. Once this is achieved, it's possible to integrate both sides separately, allowing for the solution of \( y \) in terms of \( x \).

In the problem we're exploring, we divided by \( v^2 \) to separate the variables \( v \) and \( y \). After separation, we could integrate each side with respect to its own variable. It's a bit like untangling a knot: if you can successfully isolate the strands, it's much easier to straighten them out. Similarly, if you can isolate the variables in a differential equation, it can simplify the path to finding a solution. Variable separation can be one of the first techniques attempted when faced with a first-order differential equation, as it can make a seemingly daunting problem much more manageable.