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(a) Solve the differential equation $$\frac{d y}{d x}=\frac{4 x}{y^{2}}$$ Write the solution \(y\) as an explicit function of \(x\) (b) Find the particular solution for each initial condition below and graph the three solutions on the same coordinate plane. $$y(0)=1, \quad y(0)=2, \quad y(0)=3$$

Short Answer

Expert verified
The general solution is \( y = \sqrt[3]{6x^2 + C} \), and particular solutions are \( y = \sqrt[3]{6x^2 + 1} \), \( y = \sqrt[3]{6x^2 + 8} \), \( y = \sqrt[3]{6x^2 + 9} \).

Step by step solution

01

Separate Variables

Start by separating the variables in the differential equation \( \frac{d y}{d x}=\frac{4 x}{y^{2}} \). We need to get all \( y \) terms on one side and all \( x \) terms on the other. Multiply both sides by \( y^2 \) and \( dx \) to obtain \( y^2 dy = 4x dx \).
02

Integrate Both Sides

Integrate both sides with respect to their respective variables. For the left-hand side, integrate \( \int y^2 dy \) to get \( \frac{y^3}{3} + C_1 \). For the right-hand side, integrate \( \int 4x dx \) to obtain \( 2x^2 + C_2 \).
03

Combine Integration Constants

Combine the constants of integration from both sides. Setting \( C = C_2 - C_1 \), you have the equation \( \frac{y^3}{3} = 2x^2 + C \).
04

Solve for \( y \)

Multiply both sides by 3 to solve for \( y^3 \): \( y^3 = 6x^2 + 3C \). Then solve for \( y \) to get \( y = \sqrt[3]{6x^2 + 3C} \).
05

Apply Initial Conditions

Use the initial conditions to find the particular solutions. For \( y(0)=1 \), set \( 1 = \sqrt[3]{0 + 3C} \) which gives \( C = \frac{1}{3} \). For \( y(0)=2 \), \( 2 = \sqrt[3]{0 + 3C} \) results in \( C = \frac{8}{3} \). And for \( y(0)=3 \), \( 3 = \sqrt[3]{0 + 3C} \) gives \( C = 3 \).
06

Write Particular Solutions

With the values of \( C \) found, express the particular solutions: \( y = \sqrt[3]{6x^2 + 1} \) for \( y(0)=1 \), \( y = \sqrt[3]{6x^2 + 8} \) for \( y(0)=2 \), and \( y = \sqrt[3]{6x^2 + 9} \) for \( y(0)=3 \).
07

Graph Solutions

Graph the particular solutions on the same coordinate plane to visualize how they behave for different initial conditions. Each function \( y = \sqrt[3]{6x^2 + C} \) will have its distinct curve based on the initial condition values for \( C \).

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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 fundamental technique used in solving differential equations. Essentially, it involves rearranging the equation so that all the terms involving one variable are on one side, and all the terms involving the other variable are on the opposite side.
In the exercise provided, the differential equation given is \( \frac{d y}{d x}=\frac{4 x}{y^{2}} \). To separate variables, we multiply through by \( y^2 \) and \( dx \), yielding \( y^2 dy = 4x dx \).
This step is crucial because it sets the stage for independent integration with respect to each variable in the next step. By isolating each variable, the differential equation becomes more manageable, leading to a clearer path towards finding the solution.
While separation of variables aligns terms logically, it only works on differential equations that are specifically separable. Thus, recognizing when this technique is applicable is key to correctly applying it.
Initial Conditions
Initial conditions are specific values of a solution at a given point that help us determine particular solutions to differential equations. They are essential because they allow us to find the exact solution from a family of solutions derived in the integration process.
In this exercise, we have three initial conditions: \( y(0) = 1 \), \( y(0) = 2 \), and \( y(0) = 3 \). Each of these corresponds to a unique particular solution.
Applying the initial conditions involves substituting the \( x \) and \( y \) values into the general solution and solving for the constant of integration, \( C \). This process tailors the general solution to fit the conditions specified, allowing us to express \( y \) as explicit functions for each initial scenario.
  • For \( y(0) = 1 \), the calculation provided \( C = \frac{1}{3} \).
  • For \( y(0) = 2 \), \( C = \frac{8}{3} \).
  • For \( y(0) = 3 \), \( C = 3 \).
By accurately incorporating initial conditions, we ensure the derived solution accurately reflects the specific problem scenario.
Integration
Integration is a pivotal process in solving differential equations, allowing us to reverse the differentiation process and discover functions whose derivatives meet the given equation.
For the differential equation \( y^2 dy = 4x dx \), integrating both sides is the next step after separating variables.
- For the y-component: \( \int y^2 dy = \frac{y^3}{3} + C_1 \). - For the x-component: \( \int 4x dx = 2x^2 + C_2 \).
Post integration, the combination of these expressions gives \( \frac{y^3}{3} = 2x^2 + C \), where \( C \) is a constant that incorporates \( C_1 \) and \( C_2 \).
Integration thus provides the necessary mathematical framework to move from a rate of change (derivative) to a more explanatory form (function). This step is not only about calculating antiderivatives but also about understanding how the continuous accumulation or reduction of values is represented in function form.
Graphing Solutions
Graphing solutions is an effective way to visualize and compare the behavior of differential equations under different initial conditions. After deriving the particular solutions for each case (\( y(0)=1 \), \( y(0)=2 \), and \( y(0)=3 \)), these solutions can be graphed to observe their distinctive paths.
Each particular solution, like \( y = \sqrt[3]{6x^2 + C} \), represents a different trajectory on the coordinate plane. Changes in the constant \( C \), as determined by initial conditions, shift and shape the graph of \( y \) over the x-axis.
Graphing these can:
  • Highlight how varying initial conditions affect the function's curvature and position.
  • Offer insights into the behavior of solutions over different ranges of \( x \).
  • Provide a clear visual tool to compare and contrast how each solution evolves.
Overall, graphing is not only a tool for verification but also enriches understanding by depicting solutions in a directly observable form.

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Most popular questions from this chapter

Warfarin is a drug used as an anticoagulant. After administration of the drug is stopped, the quantity remaining in a patient's body decreases at a rate proportional to the quantity remaining. The half-life of warfarin in the body is 37 hours. (a) Sketch the quantity, \(Q,\) of warfarin in a patient's body as a function of the time, \(t,\) since stopping administration of the drug. Mark the 37 hours on your graph. (b) Write a differential equation satisfied by \(Q\) (c) How many days does it take for the drug level in the body to be reduced to \(25 \%\) of the original level?

When people smoke, carbon monoxide is released into the air. In a room of volume \(60 \mathrm{m}^{3},\) air containing \(5 \%\) carbon monoxide is introduced at a rate of \(0.002 \mathrm{m}^{3} / \mathrm{min}\) (This means that \(5 \%\) of the volume of the incoming air is carbon monoxide.) The carbon monoxide mixes immediately with the rest of the air, and the mixture leaves the room at the same rate as it enters. (a) Write a differential equation for \(c(t),\) the concentration of carbon monoxide at time \(t,\) in minutes. (b) Solve the differential equation, assuming there is no carbon monoxide in the room initially. (c) What happens to the value of \(c(t)\) in the long run?

Are the statements in Problems \(59-62\) true or false? Give an explanation for your answer. The differential equation \(d y / d x=x+y\) can be solved by separation of variables.

Four pairs of species are given, with descriptions of how they interact. I. Bees/flowers: each needs the other to survive II. Owls/trees: owls need trees but trees are indifferent III. Elk/buffalo: in competition and would do fine alone IV. Fox/hare: fox eats the hare and needs it to survive Match each system of differential equations with a species pair, and indicate which species is \(x\) and which is \(y.\) (a) \(\frac{d x}{d t}=-0.2 x+0.03 x y\) \(\frac{d y}{d t}=0.4 y-0.08 x y\) (b) \(\frac{d x}{d t}=0.18 x\) \(\frac{d y}{d t}=-0.4 y+0.3 x y\) (c) \(\frac{d x}{d t}=-0.6 x+0.18 x y\) \(\frac{d y}{d t}=-0.1 y+0.09 x y\) (d) Write a possible system of differential equations for the species pair that does not have a match.

A drug is administered intravenously at a constant rate of \(r\) mg/hour and is excreted at a rate proportional to the quantity present, with constant of proportionality \(\alpha>0\) (a) Solve a differential equation for the quantity, \(Q,\) in milligrams, of the drug in the body at time \(t\) hours. Assume there is no drug in the body initially. Your answer will contain \(r\) and \(\alpha .\) Graph \(Q\) against \(t\) What is \(Q_{\infty},\) the limiting long-run value of \(Q ?\) (b) What effect does doubling \(r\) have on \(Q_{\infty} ?\) What effect does doubling \(r\) have on the time to reach half the limiting value, \(\frac{1}{2} Q_{\infty} ?\) (c) What effect does doubling \(\alpha\) have on \(Q_{\infty} ?\) On the time to reach \(\frac{1}{2} Q_{\infty} ?\)

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