91Ó°ÊÓ

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. $$ y^{\prime \prime}+\left(y^{\prime}\right)^{2}=2 e^{-y} $$ $$ \begin{array}{l}{\text { Hint: In Problem 39 the transformed equation is a Bemoulli equation. See Problem 27 in }} \\ {\text { Section 2.4. }}\end{array} $$

Step by step solution

01

Rewrite the equation using v = y'

Let \(v = y^{\prime}\). The given equation is: $$y^{\prime\prime}+\left(y^{\prime}\right)^{2}=2e^{-y}.$$ Now, replacing \(y^{\prime}\) with \(v\), we get: $$y^{\prime\prime} + v^{2} = 2e^{-y}.$$

02

Use the chain rule to reformulate the equation

Using the chain rule, \(d v / d t = (d v / d y)(d y / d t) = v(d v / d y)\). Now, replace \(y^{\prime\prime}\) in the above equation with \(v(dv/dy)\): $$v(dv/dy) + v^{2} = 2e^{-y}.$$

03

Solve the first-order differential equation

Rearrange the equation to separate variables: $$(dv/dy) = (2e^{-y}/v) - v.$$ This is a first-order differential equation that we can solve using the integrating factor method or other applicable methods. Integrating both sides, we find: $$\int v dv = \int (2e^{-y} - v^{2})dy.$$ The left side yields: $$\frac{1}{2}v^{2} + C_{1}.$$ The right side yields: $$2e^{-y} - \frac{1}{3}v^{3} + C_{2}.$$

04

Solve for v(y) and y(t)

Combining the results from the integrations, we have: $$\frac{1}{2}v^{2} + C_{1} = 2e^{-y} - \frac{1}{3}v^{3} + C_{2}.$$ From this equation, we can solve for \(v(y)\) and then \(y(t)\). Note that there will be two arbitrary constants in the final solution. So, the solution of the given second-order differential equation can be found by following these steps.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Variable Missing

In some differential equations, the independent variable, often denoted as \( t \), does not appear explicitly. This means the equation is structured such that the changes within the system do not directly rely on \( t \) but on the dependent variable \( y \) and possibly its derivatives.

When dealing with such equations, identifying that the independent variable is missing is crucial. It often allows us to reformulate the problem in a way that leverages these characteristics. This is achieved by considering another derivative or reformulating the problem with respect to the dependent variable, \( y \).

For example, if you have a second order differential equation like \( y'' = f(y, y') \), one approach is to substitute \( v = y' \), transforming it into a context where \( t \) no longer appears. This can significantly simplify the problem, making it easier to find a solution.

Chain Rule

The chain rule from calculus is a vital tool to tackle equations where variables are interdependent, such as when \( y \) depends indirectly on \( t \) through its derivatives. The chain rule helps in differentiating composite functions.

In differential equations, particularly those missing an independent variable, it helps transform the equation into a more solvable form. For instance, given \( dv/dt \), applying the chain rule allows us to write it as \((dv/dy)(dy/dt)\), which can be expressed as \( v(dv/dy) \) if \( v = y' \).

This step is crucial in rewriting the second-order differential equation so that it turns into a first-order equation in the new variable \( v \), with \( v \) being a function of \( y \). Understanding and mastering the chain rule can greatly simplify the understanding and solving of differential equations.

First Order Differential Equations

Once you identify and apply the chain rule in removing the independent variable, the goal is often to transform the original second-order equation into a first-order differential equation.

First-order differential equations are those that involve the first derivative of the function but no higher derivatives. They are generally easier to solve than second-order ones.

By converting the original equation into a first-order context, you reduce the complexity and often make use of simpler techniques for solving it. In our example, using \( v = y' \), we derived a first-order differential equation \( (dv/dy) = (2e^{-y}/v) - v \). Solving this yields a function \( v(y) \) which further aids in discovering the relation between \( y \) and \( t \).

Variable Separation Method

The variable separation method is a technique for solving differential equations by rearranging them into a form where each variable and its differential are isolated on opposite sides of the equation.

This is particularly useful for first-order differential equations. Once separated, both sides can be integrated easily with respect to their variables.

In the context of our example, after using the chain rule to reformat the equation, we achieved a form where variables could be separated: \( (dv/dy) = (2e^{-y}/v) - v \). The next step was to integrate each side, which helps find an expression for \( v \) as a function of \( y \): \( \int v dv \equals \int (2e^{-y} - v^{2})dy \).

Mastering the variable separation method can significantly bolster your ability to solve these equations, making complex problems appear much more straightforward.