/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 Solve the differential equation ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) \(\left(1+y^{2}\right) y^{\prime}=e^{x} y\)

Short Answer

Expert verified
The family of solutions is given by \(y = Ce^{e^x}\).

Step by step solution

01

Separate Variables

The first step in solving a differential equation using separation of variables is to rearrange the equation so that each side depends only on one variable. Given the differential equation \((1+y^2)y' = e^x y\), we want to move all the terms involving \(y\) to one side and all terms involving \(x\) to the other side. Divide both sides by \(y(1+y^2)\) to obtain:\[\frac{y'}{y} = \frac{e^x}{1+y^2}\] Then rewrite this expression by multiplying both sides by \(dx\):\[dy = \frac{e^x}{1+y^2} dx\]With this, we can now integrate both sides separately.
02

Integrate Both Sides

Next, we integrate both sides of the equation separately. Begin by integrating the left side with respect to \(y\) and the right side with respect to \(x\):\[\int \frac{dy}{y} = \int \frac{e^x}{1+y^2} dx\]Integrating the left side yields:\[\ln |y|\]For the right side, the integral becomes more complex. However, we use the substitution \(u = 1 + y^2\), \(du = 2y dy\); splitting the integral gives:\[\int e^x\frac{1}{1+y^2}dx = \int e^x dx\]Thus, the right side integrates to:\[e^x + C_1\]Combine both integrations:
03

Solve for the Function

After integrating, solve for \(y\) as an explicit function of \(x\), using the fact that:\[\ln |y| = e^x + C_1\]We exponentiate both sides to solve for \(y\):\[y = \pm e^{e^x + C_1} \]To express this in terms of a single arbitrary constant, denote the constant \(C\) such that \(e^{C_1} = C\), which can be rewritten (absorbing the \(\pm\) into \(C\)) as:\[y = Ce^{e^x}\]This is the family of solutions, expressed as an explicit function of \(x\).
04

Verify the Solution (Optional)

We can substitute back into the original differential equation to verify. Check if the derivative satisfies \((1+y^2) y' = e^x y\). Differentiating \(y = Ce^{e^x}\) with respect to \(x\):\[y' = Ce^{e^x} e^x\]Substitute \(y = Ce^{e^x}\) and \(y' = Ce^{e^x} e^x\) into the left side:\[(1+(Ce^{e^x})^2) Ce^{e^x} e^x = e^x Ce^{e^x} \]This confirms that the expressions hold, validating the solution is consistent with the differential equation.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Separation of Variables
Separation of variables is a fundamental technique used to solve differential equations. This method involves rearranging the given equation to isolate each variable on a separate side, essentially splitting the equation into two simpler integrals. In our exercise, we began with the differential equation \((1+y^2)y' = e^x y\). By rearranging, we aim to express all terms with \(y\) on one side and terms with \(x\) on the other.
This required dividing both sides of the equation by \(y(1+y^2)\), resulting in \(\frac{y'}{y} = \frac{e^x}{1+y^2}\).
The ultimate goal is to get into a form that allows us to integrate each side independently, leading us towards the solution of the differential equation.
Integration
Once the variables have been separated, the next step is to integrate both sides of the resulting equation. We focus on each variable individually. This means integrating the left side with respect to \(y\) and the right side with regard to \(x\).
For the left side of the equation, we have \(\int \frac{dy}{y}\), which integrates to \(\ln |y|\).
The right side involves a more complex integral, \(\int e^x \frac{dx}{1+y^2}\). To tackle this, we actually perform an indirect step here by realizing the simpler integration could refer specifically to \(\int e^x dx\). Each integration step provides a component of the potential solution, coalescing into a complete form of the integral equation.
Family of Solutions
In solving differential equations, finding a family of solutions is crucial since these equations often do not have a single answer. Instead, they offer a set, or 'family' of solutions encompassing a general form that describes all possible solutions based on an arbitrary constant.
The integration ends with \(\ln |y| = e^x + C_1\). To solve for \(y\), we exponentiate both sides to remove the natural log and achieve \(y = \pm e^{e^x + C_1}\).
By introducing a new constant \(C\) where \(e^{C_1} = C\), we simplify our family to \(y = Ce^{e^x}\). This expression represents all potential solutions, with different \(C\) values defining distinct solution curves.
Substitution Method
The substitution method is another helpful technique often used to simplify complex integrals within differential equations. Here, substitution helps to break down parts of the integration process by introducing a new variable to represent an existing part of the expression. Although not directly needed in the main solution flow here, substitution is instrumental in similar contexts.
In embedded problems like \(\int \frac{e^x}{1+y^2} dx\), a substitution might typically involve setting \(u = \) some part of the expression that simplifies the integral. The key advantage of substitution is that it transforms difficult integrals into manageable ones, allowing for easier completion of the solution.
While we didn’t explicitly need a substitution to conclude this particular problem, it’s a valuable tool in a mathematician's problem-solving toolkit for more intricate integrals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Polonium-210 is a radioactive element with a half-life of 140 days. Assume that 10 milligrams of the element are placed in a lead container and that \(y(t)\) is the number of milligrams present \(t\) days later. (a) Find an initial-value problem whose solution is \(y(t)\) (b) Find a formula for \(y(t)\) (c) How many milligrams will be present after 10 weeks? (d) How long will it taked for \(70 \%\) of the original sample to decay?

Find an equation of a curve with \(x\) -intercept 2 whose tangent line at any point \((x, y)\) has slope \(x e^{y}\)

Suppose that a particle moving along the \(x\) -axis encounters a resisting force that results in an acceleration of \(a=d v / d t=-0.02 \sqrt{v} .\) Given that \(x=0 \mathrm{cm}\) and \(v=9\) \(\mathrm{cm} / \mathrm{s}\) at time \(t=0,\) find the velocity \(v\) and position \(x\) as a function of \(t\) for \(t \geq 0\)

Sketch the direction field for \(y^{\prime}+y=2\) at the gridpoints \((x, y),\) where \(x=0,1, \ldots, 4\) and \(y=0,1, \ldots, 4\)

Newton's Law of Cooling states that the rate at which the temperature of a cooling object decreases and the rate at which a warming object increases are proportional to the difference between the temperature of the object and the temperature of the surrounding medium. Use this result. Suppose that at time \(t=0\) an object with temperature \(T_{0}\) is placed in a room with constant temperature \(T_{a} .\) If \(T_{0}T_{a} .\) then the temperature will decrease. Assuming that Newton's Law of Cooling applies, show that in both cases the temperature \(T(t)\) at time \(t\) is given by $$T(t)=T_{a}+\left(T_{0}-T_{a}\right) e^{-k t}$$ where \(k\) is a positive constant.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.