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Chapter 2: Problem 4

Find the general solution of the indicated differential equation. If possible, find an explicit solution. $$ y^{\prime}=\left(1+y^{2}\right) e^{x} $$

Short Answer

Expert verified
The general solution is \(\tan^{-1}(y) = e^x + C.\) Explicit solution: \(y = \tan(e^x + C).\)

Step by step solution

01

Separate Variables

The differential equation given is \(y' = (1 + y^2) e^x\). To solve this, we first separate variables. We can rewrite this as \(\frac{dy}{dx} = (1 + y^2) e^x\). Rearranging, we obtain \(\frac{dy}{1 + y^2} = e^x \, dx\). This separates the variable \(y\) on one side and \(x\) on the other.
02

Integrate Both Sides

Next, we integrate both sides of the equation. The left side becomes \(\int \frac{dy}{1 + y^2} = \tan^{-1}(y) + C_1\), where \(C_1\) is a constant. The right side is \(\int e^x \, dx = e^x + C_2\), where \(C_2\) is another constant. Thus, the integrated equation is \(\tan^{-1}(y) = e^x + C\), where \(C = C_2 - C_1\).
03

Solve for y

To find the explicit solution, we solve for \(y\). From \(\tan^{-1}(y) = e^x + C\), applying the tangent function to both sides gives \(y = \tan(e^x + C)\). This is the explicit solution to the differential equation.

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Key Concepts

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

Separation of Variables
Separation of Variables is a powerful method used to solve differential equations, where we aim to separate the different variables into each side of the equation. This allows us to integrate each side independently.

For the differential equation given in the problem, we start with \(y' = (1 + y^2) e^x\). Our goal is to separate \(y\) and \(x\).
Rearranging terms, we obtain \(\frac{dy}{1 + y^2} = e^x \, dx\).
  • Notice that all terms involving \(y\) are on the left side with \(dy\).
  • All terms involving \(x\) are on the right side with \(dx\).
This clear separation allows each integral to be solved individually, leading to further simplification towards a solution.
Integration
Once we have separated the variables, the next step is to integrate each side. Integration helps to find the antiderivative of a function, which is a crucial step in solving differential equations.

On the left-hand side, we integrate \(\int \frac{dy}{1 + y^2}\). This is a standard integral resulting in \(\tan^{-1}(y)\).

On the right-hand side, we have \(\int e^x \, dx\), yielding \(e^x\) as its result.

After integrating both sides, we add a constant of integration, \(C\), to account for any initial conditions that could alter the exact solution.
Our equation after integration becomes \(\tan^{-1}(y) = e^x + C\).
  • Integration transforms the problem from a differential form to an algebraic form.
  • This step is crucial for finding explicit solutions.
Explicit Solution
An explicit solution provides a direct expression for the dependent variable, which, in this exercise, is \(y\).

We start with the equation \(\tan^{-1}(y) = e^x + C\) obtained from the integration step. To solve for \(y\), apply the tangent function to both sides, leading to \(y = \tan(e^x + C)\).

This new form is the explicit solution of the original differential equation.
  • Having an explicit solution is highly beneficial for understanding the behavior of \(y\) as \(x\) changes.
  • It allows direct computation of \(y\) for any given \(x\).

Understanding explicit solutions is crucial because it often provides insight into the nature of the solution, revealing properties such as growth, oscillations, and potential singularities, which are not as apparent when left in implicit form.

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Most popular questions from this chapter

An autonomous differential equation is given in the form \(y^{\prime}=f(y)\). Perform each of the following tasks without the aid of technology. (i) Sketch a graph of \(f(y)\). (ii) Use the graph of \(f\) to develop a phase line for the autonomous equation. Classify each equilibrium point as either unstable or asymptotically stable. (iii) Sketch the equilibrium solutions in the \(t y\)-plane. These equilibrium solutions divide the ty-plane into regions. Sketch at least one solution trajectory in each of these regions. $$ y^{\prime}=9 y-y^{3} $$

Consider the initial value problem $$ y^{\prime}=3 y^{2 / 3}, \quad y(0)=0 . $$ It is not difficult to construct an infinite number of solutions. Consider $$ y(t)= \begin{cases}0, & \text { if } t \leq t_{0}, \\\ \left(t-t_{0}\right)^{3}, & \text { if } t>t_{0},\end{cases} $$ where \(t_{0}\) is any positive number. It is easy to calculate the derivative of \(y(t)\), when \(t \neq t_{0}\), $$ y^{\prime}(t)= \begin{cases}0, & \text { if } tt_{0},\end{cases} $$ but the derivative at \(t_{0}\) remains uncertain. (a) Evaluate both $$ y^{\prime}\left(t_{0}^{+}\right)=\lim _{t \backslash t_{0}} \frac{y(t)-y\left(t_{0}\right)}{t-t_{0}} $$ and $$ y^{\prime}\left(t_{0}^{-}\right)=\lim _{t / t_{0}} \frac{y(t)-y\left(t_{0}\right)}{t-t_{0}}, $$ showing that $$ y^{\prime}(t)= \begin{cases}0, & \text { if } t \leq t_{0}, \\\ 3\left(t-t_{0}\right)^{2}, & \text { if } t>t_{0} .\end{cases} $$ (b) Finally, show that \(y(t)\) is a solution of 21 . Why doesn't this example contradict Theorem 7.16?

An electric circuit, consisting of a capacitor, resistor, and an electromotive force can be modeled by the differential equation $$ R \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ where \(R\) and \(C\) are constants (resistance and capacitance) and \(q=q(t)\) is the amount of charge on the capacitor at time \(t\). For simplicity in the following analysis, let \(R=C=1\), forming the differential equation \(d q / d t+q=E(t)\). In Exercises 17-20, an electromotive force is given in piecewise form, a favorite among engineers. Assume that the initial charge on the capacitor is zero \([q(0)=0]\). (i) Use a numerical solver to draw a graph of the charge on the capacitor during the time interval \([0,4]\). (ii) Find an explicit solution and use the formula to determine the charge on the capacitor at the end of the four-second time period. E(t)= \begin{cases}5, & \text { if } 0

Use a numerical solver to sketch the solution of the given initial value problem. (i) Where does your solver experience difficulty? Why? Use the image of your solution to estimate the interval of existence. (ii) For 11-14 only, find an explicit solution; then use your formula to determine the interval of existence. How does it compare with the approximation found in part (i)? $$ \frac{d y}{d t}=\frac{t}{y+1}, \quad y(2)=0 $$

Calculate the differential \(d F\) for the given function \(F\). $$ F(x, y)=x y+\tan ^{-1}(y / x) $$
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