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In Problems, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=\frac{x+1}{y}, \text { with } y(0)=2 $$

Short Answer

Expert verified
The solution to the differential equation is \( y^2 = x^2 + 2x + 4 \).

Step by step solution

01

Separate Variables

To solve the differential equation \( \frac{d y}{d x} = \frac{x+1}{y} \), first separate the variables by multiplying both sides by \( y \, dx \). This yields: \( y \, dy = (x+1) \, dx \).
02

Integrate Both Sides

Integrate both sides of the equation separately. The left-hand side becomes \( \int y \, dy \) and the right-hand side becomes \( \int (x+1) \, dx \).
03

Solve Integrals

Evaluate the integrals: \( \int y \, dy = \frac{y^2}{2} + C_1 \) and \( \int (x+1) \, dx = \frac{x^2}{2} + x + C_2 \).
04

Combine & Simplify

Combine the results: \( \frac{y^2}{2} = \frac{x^2}{2} + x + C \) where \( C = C_2 - C_1 \). Multiply the entire equation by 2 to clear the fraction, obtaining: \( y^2 = x^2 + 2x + 2C \).
05

Apply Initial Condition

Apply the initial condition \( y(0) = 2 \) to find \( C \). Substitute \( x = 0 \) and \( y = 2 \) into \( y^2 = x^2 + 2x + 2C \) to obtain: \( 4 = 0^2 + 2(0) + 2C \).
06

Solve for the Constant

From the equation \( 4 = 2C \), solve for \( C \) to get \( C = 2 \).
07

Final Solution

Substitute \( C = 2 \) back into the equation \( y^2 = x^2 + 2x + 2C \), resulting in the final solution: \( y^2 = x^2 + 2x + 4 \).

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Key Concepts

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

Initial Condition
In differential equations, an initial condition helps us determine the specific solution that fits a given scenario. Considering any family of solutions to a differential equation, the initial condition allows us to pinpoint one particular solution among them.
In our exercise, the initial condition is given as \( y(0) = 2 \). This means when \( x = 0 \), \( y \) must equal 2.
To apply the initial condition, we plug the values into the solution after separating and integrating the equation. By substituting \( x = 0 \) and \( y = 2 \) into the developed equation, we find the constant \( C \), which allows us to adjust our solution to fit the initial condition. This step ensures that our solution not only solves the general differential equation but also fits the specific situation given.
Separation of Variables
Separation of variables is a method used to solve differential equations by rearranging terms so that each variable appears on a different side of the equation. This approach is applicable specifically to equations that can be manipulated into a form where one side of the equation involves only one variable and its derivative.
Let's illustrate how this applies to our example: Starting with \( \frac{d y}{d x} = \frac{x+1}{y} \), the strategy lies in isolating \( dy \) with \( y \) and \( dx \) with \( x \). We achieve this by multiplying both sides of the equation by \( y \, dx \), leading to \( y \, dy = (x+1) \, dx \).
Thus,
  • Left side solely involves \( y \) and \( dy \)
  • Right side solely involves \( x \) and \( dx \)
Successfully applying this technique simplifies the integration step, as the separate integrals can then be handled independently.
Integration
Integration is a fundamental step in resolving the rearranged differential equation. By computing the integral of each side independently, one can decipher the relationship between the variables.
For the separated equation \( y \, dy = (x+1) \, dx \), we integrate both sides:
  • On the left, \( \int y \, dy \) integrates to \( \frac{y^2}{2} + C_1 \)
  • On the right, \( \int (x+1) \, dx \) integrates to \( \frac{x^2}{2} + x + C_2 \)
Combining these results gives us \( \frac{y^2}{2} = \frac{x^2}{2} + x + C \), where \( C = C_2 - C_1 \). This simplification indicates that integration offers us a general solution.
Upon substituting the initial condition, integration also aids in determining the exact form of this solution by solving for the constant \( C \) and ensuring the equation suits specific initial values. This process underscores the utility of integration in solving differential equations and its role in marrying the theoretical with the practical.

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Most popular questions from this chapter

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d y}{d x}=\frac{1}{y^{3}}-\frac{1}{y} \quad, \quad y>0 $$

By breaking down each equation into two parts that you can sketch, determine how many equilib\mathrm{\\{} r i a ~ e a c h ~ d i f f e r e n t i a l ~ e q u a t i o n ~ h a s , ~ a n d ~ c l a s s i f y ~ t h e m ~ a s ~ s t a b l e ~ or unstable. You do not need to determine the location of the equilibria. $$ \frac{d x}{d t}=\frac{1}{2}-\frac{x^{2}}{x^{2}+1} $$

In Problems 13 and 14 we assumed that the per colony extinction rate was proportional to \(p .\) This means that the per colony extinction rate goes to 0 for small \(p .\) This may not be realisticsubpopulations may still go extinct even if they are not competing among themselves. One way to model this is to say that the per colony extinction rate is a function \(m(p)\) of \(p\). In Problems 15 and 16 we will assume that \(m(p)=a+b p\) for some constants \(a, b>0 .\) That is, the extinction rate increases with \(p\) because of competition between subpopulations, but \(m(p)\) does not vanish as \(p \rightarrow 0\). Then our model for proportion of occupied sites must be modified to: $$ \frac{d p}{d t}=c p(1-p)-(a+b p) p $$ where \(c, a, b\) are all positive constants. Assuming that the subpopulations obey the differential Equation (8.60) and the coefficients are \(a=1, b=2\), but \(c\) is allowed to take any value: (a) Find the equilibrium values of \(p\) (your answer will depend on the unknown coefficient \(c\) ). (b) What are the conditions on \(c\) for \(p\) to have a nontrivial equilibrium, that is, an equilibrium in which \(p \in(0,1] ?\) (c) Show that if your condition from (b) is met, then the nontrivial equilibrium is also stable.

For make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d x}{d t}=\frac{x^{2}-x}{x^{2}+1} $$

A cell constantly gains or loses small molecules to its environment because the small molecules are able to diffuse through the cell membrane. We will build a model for this process. Suppose a molecule is present in the cell at a concentration \(C(t)\), and present in its environment at a concentration \(C_{\infty}\) (you may assume \(C_{\infty}\) is a constant). One model for the diffusion of molecules across the cell membrane is that the rate at which molecules travel through the membrane is proportional to the difference in concentration between the cell and its surroundings. That is: Rate at which $$ \text { molecules flow out }=k\left(C-C_{\infty}\right) $$ of cell The constant \(k\) is known as the permeability of the membrane: \(k>0\), and \(k\) depends on the surface area of the cell and the chemistry of the membrane, as well as the type of molecule. (a) Starting with a word equation for the amount of small molecules in the cell, show, if the cell volume is \(V\), then: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right) $$ (b) Find the equilibrium of \((8.53)\) and use a graphical analysis to determine whether it is stable or unstable. (c) Suppose that the molecule we are studying is produced within the cell. The cell produces the molecule at a rate \(r\); that is, a quantity \(r\) is produced (added to the cell) in unit time. Explain why the differential equation for the concentration of molecules in the cell should be modified to: $$ \frac{d C}{d t}=-\frac{k}{V}\left(C-C_{\infty}\right)+\frac{r}{V} $$ (d) Analyze Equation (8.54) to find the equilibrium value of the cell concentration. Is this equilibrium stable or unstable? You may use a graphical argument or calculate the eigenvalue to determine the equilibrium's stability.

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