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Boyce e DiPrima - Equações Diferenciais Elementares Ed Uploaded by danielfachim Download as PDF or read online from Scribd. Flag for inappropriate. Pee ae Elementares e Problemas de Valores ao ie WILLIAM E. BOYCE RICHARD C. DIPRIMAEquacées Diferenciais Elementares e Problemas de Valores de Contorno Download as PDF or read online from Scribd. Flag for inappropriate. Veja grátis o arquivo Exercícios resolvidos Equações diferenciais elementares 9 ed Boyce e DiPrima enviado para a disciplina de Equações Diferenciais.

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Actions Shares. Embeds 0 No embeds. No notes for slide. The equilibrium solution appears topositive be , to which all other solutions converge. All solutions appear tonegative diverge away from the equilibrium solution. All solutionsnegative diverge away from www.

All solutions diverge awaynegative from the equilibrium solution. The required rates are satisfied by the differential equation.

Note that for and. Based on the direction field, for ; thus solutions with initialw values than diverge from the solution. Observe that for and. For , the slopes are all ; thus solutions with initialnegative values than diverge from the solution. All solutions appear to approach a linear asymptote. All solutions appear to converge to. First, the rate of change is small. The of these curves are at ,.

Once the solutions reach the critical value, given by the equation , the slopes become negative and negative. Applying the initial condition results in the specification ofa b! The differential equation can be rewritten asa b-. Furthermore, the convergence rate of all solutions, that is, , also increases.

In this case, the convergence rate remains the same. Consider the simpler equation. We" find that www. That is, , and hence. Invoking an initial condition ,a b a b" C! The general solution is , that is,. With , the specific solution becomes.

This solution is a: The answer in part is a general equation relating time of extinction to the value of the initial population. Since is the initial population, the appropriate: The general solution is. Solving for the rate constant, , with units ofa b per month. The hypothesis is stated mathematically as: The equation can1 gravitational constant be www. Suppose that the object is released from a height of above the ground.

Hence the amount ofmg thorium present at any time is given by.

Let be the time that it takes the isotope to decay to of its originalone-half amount. From part , it follows that , in which. The general solution of the differential equation is ,.

Taking the natural logarithm of both sides, it follows that! The differential equation governing the amount of radium is ,. The solution of the differential equation, with , isU! The rate of the chemical is. Consequently, the chemical the pond at a rate of. Since the pond is initially free of the chemical,. The differential equation can be rewritten as. Since , the"!!!! Note that. With the rate now equal to , the governing equation becomesaccumulation zero. Letting be the amount of time after the source is removed, we obtain the equation "!

It is assumed that dye is no longer entering the pool. Hence the equation that governs the amount of dye in the pool is. The initial amount of dye in the pool is. The amount of dye in the pool after four hours is obtained by setting. Let be the time that it takes to reduce the concentration level of the dye to! At that time, the amount of dye in the pool is.

Consider the differential equation. Here the parameter corresponds to the , measured in. We obtain the equation! The differential equation is second order, since the highest derivative in the equation is of order. The equation is , since the left hand side is a linear function oftwo linear C and its derivatives.

The differential equation is , since the highest derivative of the functionfourth order C is of order. The equation is also , since the terms containing the dependentfour linear variable is linear in and its derivatives. The differential equation is , since the only derivative is of order.

Thefirst order one dependent variable is , hence the equation is. The differential equation is. Furthermore, the equation is ,second order nonlinear since the dependent variable is an argument of the , which is a linearC sine function not function.

Hence the given function is a solution. Substituting into the left hand side of the equation, we find that. Hence both functions are solutions of the differential equation. Hence the function is a solution of the differential equation. The roots are ,. Then , and. The order of the partial differential equation is , since the highest derivative, intwo fact each one of the derivatives, is of.

The equation is , since the leftsecond order linear hand side is a linear function of the partial derivatives. The partial differential equation is , since the highest derivative, and infourth order fact each of the derivatives, is of order.

The equation is , since the left handfour linear side is a linear function of the partial derivatives. The partial differential equation is , since the highest derivative of thesecond order function is of order.

The equation is , due to the product on? Hence is also a solution of the differential equation.? Then the second derivatives are?

Given the function , the partial derivatives are? It follows that.! Hence is a solution of the partial differential equation.? The variable is radial distance and the angle is measured from the vertical. Rearranging the terms results in the differential equation. All slopes eventually become positive, hence all solutions will increase without bound. All solutions seem to converge to the function! Based on the direction field, the solutions eventually become oscillatory.

The differential equation can. For , the slopes are all positive, and hence the corresponding solutions increase without bound. From the resulting , thestandard form integrating factor is. The differential equation can be. All solutions eventually have positive slopes, and hence increase without bound. The integrating factor is. After multiplying both sides by , the. Multiplying both sides by ,. The integrating factor is , and the differential equation can be written as.

After writing the equation in standard orm0 , we find that the integrating factor is. From the direction field, the critical value of the initial condition seems to be. For , solutions decrease without bound. The general solution of the differential. The solutions increase or decrease, depending on the initial value. For , the solution is!