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Chapter 9: Problem 45

Solve each differential equation and initial condition and verify that your answer satisfies both the differential equation and the initial condition. $$\left\\{\begin{array}{l} y^{\prime}=2 \sqrt{y} \\ y(1)=4 \end{array}\right.$$

Short Answer

Expert verified
The solution is \( y = (x+1)^2 \), and it satisfies both the differential equation and the initial condition.

Step by step solution

01

Separation of Variables

Given the differential equation \( y' = 2\sqrt{y} \), we want to separate the variables to integrate. Rewrite the equation as \( \frac{dy}{dx} = 2\sqrt{y} \). Divide both sides by \( \sqrt{y} \) to get \( \frac{dy}{\sqrt{y}} = 2dx \).
02

Integrate Both Sides

Integrate the left side \( \int \frac{dy}{\sqrt{y}} \) and the right side \( \int 2dx \). The left side becomes \( \int y^{-1/2} dy = 2y^{1/2} + C_1 \) and the right side becomes \( 2x + C_2 \). Thus, the integrated form is \( 2y^{1/2} = 2x + C \), where \( C = C_2 - C_1 \).
03

Solve for \( y \)

To solve for \( y \), divide the entire equation \( 2y^{1/2} = 2x + C \) by 2, getting \( y^{1/2} = x + \frac{C}{2} \). Square both sides to isolate \( y \): \( y = (x + \frac{C}{2})^2 \).
04

Apply Initial Condition

Use the initial condition \( y(1) = 4 \) to find \( C \). Substitute \( x = 1 \) and \( y = 4 \) into \( y = (x + \frac{C}{2})^2 \), which gives \( 4 = (1 + \frac{C}{2})^2 \). Solving for \( C \), first take the square root \( 2 = 1 + \frac{C}{2} \) which leads to \( \frac{C}{2} = 1 \). Thus, \( C = 2 \).
05

Verify Solution

Substitute \( C = 2 \) back into the expression for \( y \): \( y = (x + 1)^2 \). Verify by differentiating \( y \) to check it satisfies the original equation. Calculating \( y' = 2(x + 1) \), replace \( y \) in \( y' = 2\sqrt{y} \) and check: \( 2(x+1) = 2\sqrt{(x+1)^2} = 2(x+1) \), which holds true. Initial condition: \( x = 1, y = 4 \) holds since \( y = (1+1)^2 = 4 \).

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Key Concepts

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

Initial Conditions
When solving a differential equation, initial conditions are crucial. They allow us to find a specific solution out of a family of solutions. In our problem, the differential equation is paired with the initial condition \( y(1) = 4 \).
This initial condition tells us exactly what the value of \( y \) is when \( x = 1 \). By substituting these values into our general solution, \( y = (x + \frac{C}{2})^2 \), we can solve for the constant \( C \).
  • Initial conditions usually come in the form \( y(x_0) = y_0 \), where \( x_0 \) is the initial value of \( x \), and \( y_0 \) is the corresponding value of \( y \).
  • Using the initial conditions allows us to tailor the solution to fit a specific scenario or data point, ensuring the solution is not general but precise to the given situation.
Separation of Variables
Separation of variables is a powerful method to solve differential equations. It works well for equations where the variables can be rewritten as a product or quotient. For our problem, the equation \( y' = 2\sqrt{y} \) is ideal for this method.
First, we rewrite the differential equation in the form \( \frac{dy}{dx} = 2\sqrt{y} \). The trick is to get all terms involving \( y \) on one side and all terms involving \( x \) on another. Thus, we manipulate the equation to get \( \frac{dy}{\sqrt{y}} = 2dx \). This sets us up perfectly for the next step: integration.
  • Always check if the equation can be separated before applying this technique. Not all differential equations can be solved in this manner.
  • After separating the variables, integration becomes simple because you treat each side of the equation as an independent function.
Integration
Once we have separated the variables in a differential equation, the next natural step is integration. In our case, we integrate both sides of \( \frac{dy}{\sqrt{y}} = 2dx \) independently.
The left side, \( \int \frac{dy}{\sqrt{y}} \), becomes \( 2y^{1/2} + C_1 \) after integration, while the right side \( \int 2dx \) results in \( 2x + C_2 \). Combining constants, we simplify to \( 2y^{1/2} = 2x + C \).
  • Integration can often introduce an arbitrary constant \( C \), representing the family of solutions to the differential equation.
  • It's essential to solve these integrals accurately as any small mistake here will affect subsequent steps and results.
    Sometimes, integrating both sides might require practice on substitution or identifying standard integral forms to simplify the process.

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Most popular questions from this chapter

A medical examiner called to the scene of a murder will usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If \(y(t)\) is the temperature (in degrees Fahrenheit) of the body \(t\) hours after the murder, and if the room temperature is \(70^{\circ},\) then \(y\) satisfies $$ \begin{aligned} y^{\prime} &=-0.32(y-70) \\ y(0) &=98.6 \text { (body temperature initially } \left.98.6^{\circ}\right) \end{aligned}$$ a. Solve this differential equation and initial condition. b. Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the \text { body when it was discovered was } 80^{\circ} .

For each initial value problem: a. Use an Euler's method graphing calculator program to find the estimate for \(y(2)\). Use the interval [0,2] with \(n=50\) segments. b. Solve the differential equation and initial condition exactly by separating variables or using an integrating factor. c. Evaluate the solution that you found in part (b) at \(x=2\). Compare this actual value of \(y(2)\) with the estimate of \(y(2)\) that you found in part (a). $$ \begin{array}{l} \frac{d y}{d x}=-y \\ y(0)=1 \end{array} $$

For each initial value problem, calculate the Euler approximation for the solution on the interval [0,1] using \(n=4\) segments. Draw the graph of your approximation. (Carry out the calculations "by hand" with the aid of a calculator, rounding to two decimal places. Answers may differ slightly, depending on when you do the rounding.) $$ \begin{array}{l} y^{\prime}=x+2 y \\ y(0)=1 \end{array} $$

The following problems extend and augment the material presented in the text. BIOMEDICAL: Fick's Law Fick's Law governs the diffusion of a solute across a cell membrane. According to Fick's Law, the concentration \(y(t)\) of the solute inside the cell at time \(t\) satisfies \(\frac{d y}{d t}=\frac{k A}{V}\left(C_{0}-y\right),\) where \(k\) is the diffusion constant, \(A\) is the area of the cell membrane, \(V\) is the volume of the cell, and \(C_{0}\) is the concentration outside the cell. a. Find the general solution of this differential equation. (Your solution will involve the constants \(k, A, V\) and \(C_{0}\).) b. Find the particular solution that satisfies the initial condition \(y(0)=y_{0},\) where \(y_{0}\) is the initial concentration inside the cell.

For each initial value problem: a. Use an Euler's method graphing calculator program to find the estimate for \(y(2)\). Use the interval [0,2] with \(n=50\) segments. b. Solve the differential equation and initial condition exactly by separating variables or using an integrating factor. c. Evaluate the solution that you found in part (b) at \(x=2\). Compare this actual value of \(y(2)\) with the estimate of \(y(2)\) that you found in part (a). $$ \begin{array}{l} y^{\prime}=\frac{x}{y} \\ y(0)=1 \end{array} $$
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