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Solve the equation $$ \frac{d y}{d x}=\frac{a y+b}{c y+d} $$ $$ \text { where } a, b, c, \text { and } d \text { are constants. } $$

Short Answer

Expert verified
Answer: The implicit solution for y is $$\frac{1}{a}\ln(|ay+b|) = \frac{1}{c}\ln(|cy+d|) + K$$.

Step by step solution

01

Separate the Variables

To separate the variables, rearrange the equation by multiplying both sides by $$\frac{1}{ay+b}$$ and $$dx$$. This gives us: $$ \frac{1}{ay+b}\frac{dy}{dx}=1\frac{dx}{cy+d} $$
02

Integrate both sides

Now, we must integrate each fraction with respect to their respective variables. $$ \int\frac{1}{ay+b}dy=\int\frac{1}{cy+d}dx $$ Note that, because a, b, c, and d are constants, they can be considered as constant coefficients in the integration process. Therefore, we can perform substitution. Let $$u = ay+b$$, then $$\frac{du}{da} = a$$, which means $$\frac{1}{a}du = dy$$. Similarly, let $$v = cy+d$$, then $$\frac{dv}{dc} = c$$, which means $$\frac{1}{c}dv = dx$$. Now, the above equation can be rewritten as, $$ \int\frac{1}{a} \frac{1}{u}du = \int\frac{1}{c} \frac{1}{v}dv $$
03

Solve each integral

Integrate both sides of the equation: $$ \frac{1}{a}\int\frac{1}{u}du = \frac{1}{c}\int\frac{1}{v}dv $$ $$ \frac{1}{a}\ln(|u|) + K_1 = \frac{1}{c}\ln(|v|) + K_2 $$
04

Re-substitute original variables

Replace u and v with their original expressions: $$ \frac{1}{a}\ln(|ay+b|) + K_1 = \frac{1}{c}\ln(|cy+d|) + K_2 $$ We can combine $$K_1$$ and $$K_2$$ into one constant, say K: $$ \frac{1}{a}\ln(|ay+b|) = \frac{1}{c}\ln(|cy+d|) + K $$
05

Solve for y

This is the implicit solution of the given differential equation. The explicit solution for y may be difficult to obtain, but this representation provides a satisfactory result. The student can use this implicit solution to solve particular cases when initial conditions are provided.

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Key Concepts

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

Separation of Variables
The technique of separation of variables is a method used to solve first-order ordinary differential equations. It involves rearranging the equation so that each side contains only one variable—either the dependent variable (typically y) or the independent variable (typically x).

To apply separation of variables to the given equation \[\frac{d y}{d x}=\frac{a y+b}{c y+d}\], we multiply both sides by \[\frac{1}{ay+b}dx\], effectively isolating dy on one side and dx on the other. This rearrangement is crucial because it allows us to integrate both sides independently, moving us closer to finding a solution for y. It's important to note that separation of variables is not always possible, but when it is, it simplifies the process significantly.

Integrating each side with respect to its respective variable yields functions of y on one side and functions of x on the other. Once integrated, we then have an implicit solution that typically includes an integration constant.
Integrating Factor
An integrating factor is a mathematical tool that's used to solve certain types of differential equations, notably those that can't be easily separated. When separation of variables is not applicable, the method of integrating factors can often be used, especially on linear first-order ordinary differential equations. While not directly applicable to the exercise provided because we could use separation of variables, understanding integrating factors is extremely valuable.

In general terms, an integrating factor is a function that multiplies both sides of a differential equation to make it easier to solve. It is typically denoted as \(\mu(x)\), where \(\mu(x)\) is a function chosen such that the left side of the equation becomes the exact derivative of a product of \(\mu(x)\) and the function we are trying to solve for.
First-Order Ordinary Differential Equations
First-order ordinary differential equations (ODEs) are equations that contain a first derivative and no higher derivatives. These equations can often describe physical phenomena, such as rate processes in physics, chemistry, and biology. The general form of a first-order ODE is \(\frac{dy}{dx} = f(x, y)\), where f is a given function of x and y.

First-order ODEs can be classified into several types, including separable, linear, and exact differential equations. There are specific methods to tackle each type, such as separation of variables for separable equations, using an integrating factor for linear non-separable equations, or finding a potential function for exact equations. The provided exercise handles a first-order ODE that was successfully solved using separation of variables, showing the diversity of strategies available for these kinds of differential equations.
Integration of Rational Functions
Integration of rational functions is a process where we integrate functions that are ratios of polynomials. In the context of differential equations, integrating rational functions often occurs after separating variables.

For the given problem, we deal with the integrals of \(\frac{1}{ay+b}\) and \(\frac{1}{cy+d}\), both of which are rational functions. A common technique for integrating rational functions is to perform a u-substitution. This strategy simplifies the integrals into a form that is easier to handle - in this case, resulting in the natural logarithm of the absolute value of a linear function.

After integration, we typically reintroduce the original variables, leading to an implicit form of the solution involving natural logs. The integration constants, \(K_1\) and \(K_2\), represent the arbitrary constants of integration that arise in the indefinite integration process, and combined, they contribute to the general solution of the differential equation.

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Most popular questions from this chapter

Sometimes it is possible to solve a nonlinear equation by making a change of the dependent variable that converts it into a linear equation. The most important such equation has the form $$ y^{\prime}+p(t) y=q(t) y^{n} $$ and is called a Bernoulli equation after Jakob Bernoulli. the given equation is a Bernoulli equation. In each case solve it by using the substitution mentioned in Problem 27(b). \(y^{\prime}=r y-k y^{2}, r>0\) and \(k>0 .\) This equation is important in population dynamics an is discussed in detail in Section 2.5 .

let \(\phi_{0}(t)=0\) and use the method of successive approximations to solve the given initial value problem. (a) Determine \(\phi_{n}(t)\) for an arbitrary value of \(n .\) (b) Plot \(\phi_{n}(t)\) for \(n=1, \ldots, 4\). Observe whether the iterates appear to be converging. (c) Express \(\lim _{n \rightarrow \infty} \phi_{n}(t)=\phi(t)\) in terms of elementary functions; that is, solve the given initial value problem. (d) Plot \(\left|\phi(t)-\phi_{n}(t)\right|\) for \(n=1, \ldots, 4 .\) For each of \(\phi_{1}(t), \ldots . \phi_{4}(t)\) estimate the interval in which it is a reasonably good approximation to the actual solution. $$ y^{\prime}=-y / 2+t, \quad y(0)=0 $$

Draw a direction field for the given differential equation and state whether you think that the solutions are converging or diverging. $$ y^{\prime}=y(3-t y) $$

Use the technique discussed in Problem 20 to show that the approximation obtained by the Euler method converges to the exact solution at any fixed point as \(h \rightarrow 0 .\) $$ y^{\prime}=\frac{1}{2}-t+2 y, \quad y(0)=1 \quad \text { Hint: } y_{1}=(1+2 h)+t_{1} / 2 $$

Suppose that a rocket is launched straight up from the surface of the earth with initial velocity \(v_{0}=\sqrt{2 g R}\), where \(R\) is the radius of the earth. Neglect air resistance. (a) Find an expression for the velocity \(v\) in terms of the distance \(x\) from the surface of the earth. (b) Find the time required for the rocket to go \(240,000\) miles (the approximate distance from the earth to the moon). Assume that \(R=4000\) miles.

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