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Chapter 1: Problem 53

Consider the differential equation $$y^{\prime}=F(a x+b y+c)$$ Where \(a, b \neq 0,\) and \(c\) are constants. Show that the change of variables from \(x, y\) to \(x, V,\) where $$V=a x+b y+c$$ reduces Equation (1.8.17) to the separable form $$ \frac{1}{b F(V)+a} d V=d x $$

Short Answer

Expert verified
Short answer: Perform the change of variables, from \(x\), \(y\) to \(x\), \(V\), with \(V = ax + by + c\), and differentiate \(V\) with respect to \(x\): $$ \frac{dV}{dx} = a + b\frac{dy}{dx}. $$ Replace \(y'\) by \(\frac{dV}{dx} - a\) in the given differential equation: $$ \frac{dV}{dx} - a = F(V). $$ Solve for \(\frac{dV}{dx}\): $$ \frac{dV}{dx} = F(V) + a. $$ Express the differential equation in the separable form: $$ \frac{1}{bF(V) + a} dV = dx. $$

Step by step solution

01

Derive an expression for \(V\)

We are given that \(V=ax+by+c\). Simply differentiating \(V\) with respect to \(x\), we get: $$ \frac{dV}{dx} = a + b\frac{dy}{dx}. $$
02

Replace \(y'\) by \(\frac{dV}{dx} - a\) in the given differential equation

We know from the given differential equation that \(y' = F(ax+by+c)\). We also have \(y' = \frac{dV}{dx} - a\). Therefore, we can write: $$ \frac{dV}{dx} - a = F(ax + by +c). $$ Since \(ax+by+c=V\), we have: $$ \frac{dV}{dx} - a = F(V). $$
03

Solve for \(\frac{dV}{dx}\)

Next, we will solve for \(\frac{dV}{dx}\). From the equation above, we can write: $$ \frac{dV}{dx} = F(V) + a. $$
04

Express the differential equation in the separable form

We can now rewrite the differential equation in the separable form by dividing both sides with \(bF(V)+a\). Thus, we have: $$ \frac{1}{bF(V) + a} dV = dx. $$ This shows that the given change of variables indeed reduces the original differential equation to the desired separable form.

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a class of differential equations that can be split into separate variables on opposite sides of the equation. Essentially, this means that all terms involving the dependent variable (for example, y) and its derivatives can be expressed on one side of the equation, while all terms involving the independent variable (generally x) reside on the other.

Mathematically, a separable differential equation can be written in the form \( \frac{dy}{dx} = g(x)h(y) \). Here, the functions g(x) and h(y) are only dependent on x and y, respectively. Solving such equations involves integrating both sides, often after rearranging the equation to isolate the variables:
  • Integrate the function with x with respect to x.
  • Integrate the function with y with respect to y.
The integration may yield a direct solution or involve an implicit relation between x and y.
Change of Variables
The change of variables is a powerful method used to simplify differential equations by substituting one set of variables with another, more manageable set. This approach can transform a complex or non-standard equation into a form that is easier to solve. In the exercise, the substitution V = ax + by + c is suggested.

When applying a change of variables, it's crucial to derive the relationship between the new and old variables correctly, particularly how derivatives transform. The differential of V with respect to x, denoted \(\frac{dV}{dx}\), is computed while keeping in mind that V is a function of x and y. If y itself is a function of x, which it typically is in differential equations, then \(\frac{dy}{dx}\) also appears in the expression for \(\frac{dV}{dx}\).
Solving Differential Equations
Solving differential equations involves finding a function or a set of functions that satisfy the equation. Solving can mean different things: sometimes it means finding a closed-form expression for the solution, and other times it may involve a numerical method or a series expansion. The goal is always to express the dependent variable as a function of the independent variable(s).

In our example, we aim to find y as a function of x, but the initial form of the differential equation does not allow a straightforward integration. By using a change of variables, we transform the problem into finding V as a function of x, which then can be related back to y. Solving this transformed equation often involves integration, and applying boundary conditions or initial values to arrive at a particular solution.
Variable Separation
Variable separation is a strategy used primarily for solving separable differential equations. This technique involves rearranging an equation so that all terms with one variable and its differential are on one side of the equation and all terms with the other variable and its differential are on the other side.

In the context of the given exercise, variable separation is achieved through the change of variables. The transformation \( y^\prime = F(ax + by + c) \) does not initially appear to be separable. But by expressing it in terms of V, and then rearranging terms, we obtain \( \frac{1}{bF(V) + a} dV = dx \), which is a separable form. Here, \( dV \) and \( dx \) are separated, allowing for direct integration on both sides:
  • Integrate \( \frac{1}{bF(V) + a} \) with respect to V.
  • Integrate 1 with respect to x.
This is the essence of variable separation and illustrates why it's a fundamental method in solving differential equations.

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Most popular questions from this chapter

Determine which of the five types of differential equations we have studied the given differential equation falls into, and use an appropriate technique to find the solution to the initial-value problem. $$\frac{d y}{d x}-(\sin x) y=e^{-\cos x}, y(0)=\frac{1}{e}$$

Let \(F_{1}\) and \(F_{2}\) be two families of curves with the property that whenever a curve from the family \(F_{1}\) intersects one from the family \(F_{2},\) it does so at an angle \(a \neq \pi / 2 .\) If we know the equation of \(F_{2},\) then it can be shown (see Problem 23 in Section 1.1 ) that the differential equation for determining \(F_{1}\) is $$ \frac{d y}{d x}=\frac{m_{2}-\tan a}{1+m_{2} \tan a} $$ where \(m_{2}\) denotes the slope of the family \(F_{2}\) at the point \((x, y)\) Use Equation \((1.8 .16)\) to determine the equation of the family of curves that cuts the given family at an angle \(\alpha=\pi / 4\) $$y=c x^{6}$$

Each spring, sandhill cranes migrate through the Platte River valley in central Nebraska. An estimated maximum of a half-million of these birds reach the region by April 1 each year. If there are only 100,000 sandhill cranes 15 days later and the sandhill cranes leave the Platte River valley at a rate proportional to the number of sandhill cranes still in the valley at the time, (a) How many sandhill cranes remain in the valley 30 days after April \(1 ?\) (b) How many sandhill cranes remain in the valley 35 days after April \(1 ?\) (c) How many days after April 1 will there be less than 1000 sandhill cranes in the valley?

Determine which of the five types of differential equations we have studied the given equation falls into (see Table \(1.12 .1),\) and use an appropriate technique to find the general solution. $$y^{\prime}+y \sin x=\sin x$$

Let \(F_{1}\) and \(F_{2}\) be two families of curves with the property that whenever a curve from the family \(F_{1}\) intersects one from the family \(F_{2},\) it does so at an angle \(a \neq \pi / 2 .\) If we know the equation of \(F_{2},\) then it can be shown (see Problem 23 in Section 1.1 ) that the differential equation for determining \(F_{1}\) is $$ \frac{d y}{d x}=\frac{m_{2}-\tan a}{1+m_{2} \tan a} $$ where \(m_{2}\) denotes the slope of the family \(F_{2}\) at the point \((x, y)\) Use Equation \((1.8 .16)\) to determine the equation of the family of curves that cuts the given family at an angle \(\alpha=\pi / 4\) $$x^{2}+y^{2}=c$$
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