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Chapter 36: Problem 850

Find \(y\) if \(y^{\prime \prime}=y^{\prime} e^{y}\) and \(y(0)=-1, y^{\prime}(0)=0\)

Short Answer

Expert verified
The solution to the given problem is expressed as an integral: \[y(y) = \int e^{e^{y} - e^{-1}} dy + K\] where K is a constant that can be found using the condition \(y(0) = -1\). Unfortunately, there is no elementary expression for the function y(y) in terms of the given differential equation, and it should be expressed or approximated using numerical methods, if needed.

Step by step solution

01

Rewrite the given differential equation into a first-order equation

First, let's rewrite the given second-order differential equation \(y^{\prime\prime}=y^{\prime} e^y\) into a first-order equation. We define a new variable \(u(y) = y^{\prime}(y)\), so that \(u^{\prime}(y) = y^{\prime\prime}(y)\). Thus, the given equation becomes \(u^{\prime}(y) = u(y) e^{y}\).
02

Solve the first-order differential equation

Now, let's solve the first-order differential equation \(u^{\prime}(y) = u(y) e^{y}\). This is a separable differential equation; we can rewrite it as \(\frac{u^{\prime}(y)}{u(y)} = e^{y}\). Integrating both sides with respect to y, we get: \[\int \frac{u^{\prime}(y)}{u(y)} dy = \int e^{y} dy .\] Applying the natural logarithm properties and reverse chain rule, we find: \[\ln| u(y) | = e^{y} + C .\] Now, exponentiate both sides to solve for \(u(y)\): \[u(y) = e^{e^{y} + C}\] or \[y^{\prime}(y) = e^{e^{y} + C} .\]
03

Use the given conditions to find the particular solution

We are given the conditions \(y(0) = -1\) and \(y^{\prime}(0) = 0\). First, let's find the value of \(C\) using \(y^{\prime}(0) = 0\). \[0 = e^{e^{0} + C}\] \[C = -1\] Now, we have the expression for the derivative of y: \[y^{\prime}(y) = e^{e^{y} - 1} .\] Next, we must find the function y(y). We do this by observing that e^(-1) equals the right-hand side of the equation when y equals 0. So we have \(y^{\prime}(y) = e^{e^{y} - e^{-1}}\). Integrating this expression with respect to y, we get: \[y(y) = \int e^{e^{y} - e^{-1}} dy .\] This integral does not have an elementary closed-form solution, so the solution for y(y) should be expressed in terms of the integral.
04

Final Answer

The solution to the given problem is expressed as an integral: \[y(y) = \int e^{e^{y} - e^{-1}} dy + K\] where K is a constant that can be found using the condition \(y(0) = -1\). Unfortunately, there is no elementary expression for the function y(y) in terms of the given differential equation, and it should be expressed or approximated using numerical methods, if needed.

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Key Concepts

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

First-Order Differential Equations
Differential equations are mathematical equations that relate a function to its derivatives. A first-order differential equation involves only the first derivative of the function and no higher derivatives.

These equations often describe various phenomena such as growth and decay, cooling processes, or simple motion. They come in different forms and involve various techniques of solution, which makes them fundamental in mathematical modeling of real-world situations.

For example, the equation \(dy/dx = f(x)\) is a first-order differential equation where \(f(x)\) is a function of x only. Solving it typically involves finding an antiderivative or using separation of variables if the equation is separable.
Separable Differential Equations
A separable differential equation is a specific type of first-order equation that can be written in the form \(N(y) dy = M(x) dx\), where \(N\) and \(M\) are functions of \(y\) and \(x\) respectively. The key characteristic that makes an equation 'separable' is that the variables can be moved to opposite sides of the equation.

To solve a separable differential equation, the following steps are taken:
  • Rearrange the equation to isolate \(dy\) and \(dx\) on opposite sides.
  • Integrate both sides, often leading to an implicit solution involving \(y\) and \(x\).
  • Solve for \(y\) if an explicit solution is desired and possible.
Estimating solutions to separable equations visually often involves sketching slope fields or using numerical methods.
Integration of Differential Equations
The integration of differential equations is a method used to find the original function from its derivative(s). While integration in calculus often involves finding antiderivatives, with differential equations, it involves a bit more strategy.

The goal is to find an unknown function that satisfies the differential equation. In the case of first-order equations, direct integration can often be used if the equation is separable. However, some differential equations may require more sophisticated techniques, such as substitution or even numerical methods when analytical solutions are difficult or impossible to find.

When integrating to solve a differential equation, constants of integration arise. Initial conditions are used to solve for these constants, providing us with the particular solution to the differential equation given specific starting values.

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Most popular questions from this chapter

Solve the initial-value problem for a non-linear spring, $$ \begin{array}{rl} x+\mathrm{kx}+\mathrm{k}^{\prime} \mathrm{x}^{3}=0 & \mathrm{x}(0)=\mathrm{c} \\\ & \mathrm{x}(0)=0, \end{array} $$ using the perturbation series method.

The two non-linear systems: $$ \begin{aligned} &(\mathrm{dx} / \mathrm{dt})=2 \mathrm{xy} ; \\ &(\mathrm{dy} / \mathrm{dt})=3 \mathrm{y}^{2}-\mathrm{x}^{2} \\ &(\mathrm{dx} / \mathrm{dt})=\mathrm{x}^{2} \\ &(\mathrm{dy} / \mathrm{dt})=2 \mathrm{y}^{2}-\mathrm{xy} . \end{aligned} $$ have the critical point \((0,0)\). Discuss the nature and stability of the critical point.

Solve the differential equation \(\mathrm{xy}(\mathrm{dy} / \mathrm{dx})^{2}+(\mathrm{x}+\mathrm{y})(\mathrm{dy} / \mathrm{dx})+1=0\)

Solve the non-linear differential equation $$ \mathrm{x}=\mathrm{y}-\mathrm{y}^{\prime 2} $$

Find all the real critical points of the nonlinear system: $$ \begin{aligned} &(\mathrm{dx} / \mathrm{dt})=8 \mathrm{x}-\mathrm{y}^{2} \\ &(\mathrm{dy} / \mathrm{dt})=-6 \mathrm{y}+6 \mathrm{x}^{2} \end{aligned} $$ and determine the type and stability of each of these critical points.
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