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Solve the following differential equations by using the method of substitution to put them into the form \(\frac{d y}{d t}=k y\). (a) \(\frac{d P}{d t}=0.3(1000-P)\) (b) \(\frac{d M}{d t}=0.4 M-2000\)

Short Answer

Expert verified
The solutions to the given differential equations are \(P = 1000 - Ce^{-0.3t}\) and \(M = 5000 + Ce^{0.4t}\), where C is the constant of integration.

Step by step solution

01

Identify the differential equations

The first differential equation is \(\frac{dP}{dt}=0.3(1000-P)\) and the second is \(\frac{dM}{dt}=0.4M-2000\).
02

Apply method of substitution for the first equation

Rearrange the terms of the first differential equation to make it in the form \(\frac{dP}{dt}=0.3P\). Then use \(y = 1000 - P\) to simplify it. The differential equation becomes \(\frac{dy}{dt} = -0.3y\). Now it's in the desired format.
03

Solve the first equation

The equation \(\frac{dy}{dt} = -0.3y\) is a simple first order linear differential equation, which solution can be found by integration. Integration yields \(y = Ce^{-0.3t}\), where C is the constant of integration, determined by the initial conditions. To find P, substitute y back into the equation, so we obtain \(P = 1000 - Ce^{-0.3t}\).
04

Apply method of substitution for the second equation

Rearrange the terms of the second differential equation to make it in the form \(\frac{dM}{dt}=0.4M\). Then use \(y = M - 5000\) to simplify it. The differential equation becomes \(\frac{dy}{dt} = 0.4y\). Now this, too, is in the desired format.
05

Solve the second equation

The equation \(\frac{dy}{dt} = 0.4y\) can be solved the way as the first one. Integration yields \(y = Ce^{0.4t}\), where C is the constant of integration. Substituting y back into the equation gives \(M = 5000 + Ce^{0.4t}\).

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Key Concepts

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

Method of Substitution
When faced with a differential equation that does not initially appear to be easily solvable, the method of substitution can be a very effective tool. This technique involves replacing a part of the original equation with a new variable to simplify the equation to a more recognizable form.

The aim is to transform the given differential equation into one for which we already know the solution method. For instance, in many cases, the goal is to produce a first order linear differential equation of the form \(\frac{dy}{dt}=ky\), where \(k\) is a constant.

By introducing a new variable, say \(y\), we essentially reframe the problem, which can then be solved using standard techniques like separation of variables or integrating factors. Once the solution for \(y\) is found, we use our substitution to express the solution in terms of the original variable.
First Order Linear Differential Equations
First order linear differential equations are a commonly encountered type of differential equation. These equations have the standard form \(\frac{dx}{dt}+p(t)x=q(t)\), where \(p(t)\) and \(q(t)\) are functions of \(t\), and the solution involves finding the function \(x(t)\).

One powerful method to solve these equations is by using an integrating factor, which is a function that, when multiplied by the equation, allows it to be written in a form that can be integrated directly. However, when the equation is already in the form of \(\frac{dx}{dt}=kx\), with \(k\) being a constant, solving it becomes even simpler as it then involves direct integration.
Integration of Functions
Integration is a fundamental concept in calculus, often thought of as the inverse operation to differentiation. It is the process of finding the area under a curve or, in other words, determining the antiderivative or indefinite integral of a function.

The general process of integrating a function involves finding a function \(F\) such that \(F'(x) = f(x)\), where \(f(x)\) is the given function to be integrated. The resulting function \(F\) includes a \(+ C\), representing the constant of integration, which is determined based on initial conditions or boundary values provided with the problem.
Constant of Integration
When we find the indefinite integral of a function, we introduce the constant of integration, usually denoted as \(C\). This constant represents the infinite number of antiderivatives that a function can have since the derivative of a constant is zero.

The value of \(C\) is determined by initial conditions or boundary conditions given with the problem. These conditions are specific values of the function or its derivatives at certain points. By applying these conditions, we can solve for the constant \(C\), providing us with the particular solution to the differential equation suited to the problem's context.

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Most popular questions from this chapter

A canister contains 10 liters of blue paint. Paint is being used at a rate of 2 liters per hour and the canister is being replenished at a rate of 2 liters per hour by a pale blue paint that is \(80 \%\) blue and \(20 \%\) white. Assuming the canister is well-mixed, write a differential equation whose solution is \(w(t)\), the amount of white paint in the canister at time \(t\). Specify the initial condition.

In the beginning of a chemical reaction there are 600 moles of substance \(\mathrm{A}\) and none of substance \(\mathrm{B}\). Over the course of the reaction, the 600 moles of substance \(\mathrm{A}\) are converted to 600 moles of substance B. (Each molecule of A is converted to a molecule of \(\mathrm{B}\) via the reaction.) Suppose the rate at which \(\mathrm{A}\) is turning into \(\mathrm{B}\) is proportional to the product of the number of moles of \(\mathrm{A}\) and the number of moles of \(\mathrm{B}\). (a) Let \(N=N(t)\) be the number of moles of substance \(\mathrm{A}\) at time \(t .\) Translate the statement above into mathematical language. (Note: The number of moles of substance B should be expressed in terms of the number of moles of substance A.) (b) Using your answer to part (a), nd \(\frac{d^{2} N}{d t^{2}}\). Your answer will involve the proportionality constant used in part (a). (c) \(N(t)\) is a decreasing function. The rate at which \(N\) is changing is a function of \(N\), the number of moles of substance A. When the rate at which \(\mathrm{A}\) is being converted to \(\mathrm{B}\) is highest, how many moles are there of substance \(\mathrm{A}\) ?

Solve the differential equations. \(y^{\prime \prime}+25 y=0\)

A lake contains \(10^{10}\) liters of water. Acid rain containing \(0.02\) milligrams of pollutant per liter of rain falls into the lake at a rate of \(10^{3}\) liters per week. An outlet stream drains away \(10^{3}\) liters of water per week. Assume that the pollutant is always evenly distributed throughout the lake, so the runoff into the stream has the same concentration of pollutant as the lake as a whole. The volume of the lake stays constant at \(10^{10}\) liters because the water lost from the runoff balances exactly the water gained from the rain. (a) Write a differential equation whose solution is \(P(t)\), the number of milligrams of pollutant in the lake as a function of \(t\) measured in weeks. (b) Find any equilibrium solutions. (c) Sketch some representative solution curves. (d) How would you alter the differential equation if there was a dry spell and rain was falling into the lake at a rate of only \(10^{2}\) liters per week.

Solve the given differential equation. \(\frac{d y}{d x}=x y^{2}\)

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