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Chapter 31: Problem 14

Sketch a representative family of solutions for each of the following differential equations. (a) \(\frac{d y}{d t}=t^{2}\) (b) \(\frac{d y}{d t}=y^{2}\)

Short Answer

Expert verified
The solution for equation (a) is \(y(t)=\frac{1}{3}t^3+C\), forming a family of cubic curves. The solution for equation (b) is \(y(t)=-\frac{1}{t + C}\), forming a family of hyperbola.

Step by step solution

01

Solving Equation (a)

To solve \(\frac{d y}{d t}=t^{2}\), integrate both sides of the equation with respect to \(t\), this gives the solution \(y(t)=\int t^2 dt = \frac{1}{3}t^3+C\), where \(C\) is the constant of integration. These represent a family of cubic curves in the \(y-t\) plane.
02

Sketching Equation (a) Solution

Sketch the family of solution curves \(y(t)=\frac{1}{3}t^3+C\). As \(C\) varies, the entire set forms a family of cubic curves. All curves will share the common property that the slope of the curve at any point \((x,y)\) on it is given by \(t^2\).
03

Solving Equation (b)

To solve the equation \(\frac{d y}{d t}=y^{2}\), separate the variables: \(\frac{1}{y^2} dy = dt\). Integrating both sides then gives: \(-\frac{1}{y} = t + C\), where \(C\) again is the constant of integration. This can be rewritten as \(y(t) = -\frac{1}{t+C}\). This represents a family of hyperbola in the \(y-t\) plane.
04

Sketching Equation (b) Solution

Sketch the family of solution curves \(y(t) = -\frac{1}{t+C}\). As \(C\) varies, the entire set forms a family of hyperbola, and all curves share the property that the slope of the tangent line at a point \((t, y)\) is given by \(y^2\).

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus, widely used to solve differential equations.Integrating involves finding a function whose derivative is the given function.This process is called anti-differentiation.
To solve the differential equation \( \frac{d y}{d t}=t^{2} \), we integrate both sides with respect to \( t \).This means calculating the integral of \( t^2 \) with respect to \( t \), which gives the solution \( y(t)=\int t^2 dt = \frac{1}{3}t^3+C \).Here, \( C \) is known as the constant of integration, representing an infinite set of solutions or a 'family' of functions.
  • Integration helps in reversing the effect of differentiation.
  • It allows us to find specific functions that satisfy certain initial or boundary conditions.
  • The constant of integration \( C \) ensures that all possible solutions to a differential equation are captured.
Family of Solutions
A family of solutions to a differential equation includes all possible solutions characterized by different values of the constant of integration \( C \).In the case of the first differential equation, the family of solutions forms cubic curves \( y(t)=\frac{1}{3}t^3 + C \).
The constant \( C \) can take any real value, causing the curves to shift vertically in the \( y-t \) plane.Similarly, for the second equation \( y(t) = -\frac{1}{t+C} \), the family of solutions are hyperbolas, where each curve in the family is defined by a different \( C \).
  • Each member of the family of solutions satisfies the original differential equation.
  • The concept emphasizes how varying initial conditions lead to different paths, visualizing a range of possibilities.
  • A clear understanding of families of solutions helps in predicting behavior in dynamic systems.
Separation of Variables
Separation of variables is a method for solving certain types of differential equations by rearranging terms.The idea is to separate the equation into two parts, each containing only one variable.
In the example \( \frac{d y}{d t}=y^{2} \), by arranging terms, we arrive at \( \frac{1}{y^2} dy = dt \).This allows us to integrate both sides separately, ultimately finding \( -\frac{1}{y} = t + C \).
  • Separation of variables is particularly useful for first-order ordinary differential equations (ODEs) that can be written in a separable form.
  • This technique simplifies equations, making them easier to solve analytically.
  • Understanding this method is key in transforming complex real-world problems into manageable mathematical forms.

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Most popular questions from this chapter

The population of wildebeest in the Serengeti was decimated by a rinderpest plague in the \(1950 \mathrm{~s}\). In 1961 the Serengeti supported a population of a quarter of a million wildebeest. By 1978 the wildebeest population was \(1.5\) million and by 1991 it had reached 2 million. (Craig Packer Into Africa, Chicago, The University of Chicago Press, 1996 p. \(250 .\) ) Given this data, would you be more inclined to model the growth of the wildebeest population using an exponential growth model or using a logistic growth model? Explain your reasoning.

Solve the differential equations. \(y^{\prime \prime}-2 y^{\prime}+y=0\)

Give an example of a differential equation with constant solutions at \(y=-1\) and \(y=4\) with the characteristics specified. (a) The equilibrium at \(y=-1\) is stable; the equilibrium at \(y=4\) is unstable. (b) The equilibrium at \(y=-1\) is unstable; the equilibrium at \(y=4\) is stable. (c) Neither equilibrium solution is stable.

(a) Find \(\lambda\) such that \(y=e^{\lambda t}\) is a solution to \(y^{\prime \prime}+4 y^{\prime}+4 y=0\). (b) Verify that \(y=t e^{\lambda t}\) is also a solution to \(y^{\prime \prime}+4 y^{\prime}+4 y+0\).

Do a qualitative analysis of the family of solutions to each of the differential equations below. Then, in another color pen or pencil, highlight the graphs of the solutions corresponding to the given initial conditions. (If the solution is asymptotic to a horizontal line, draw and label that line.) \(\begin{array}{llll}\text { (a) } \frac{d y}{d t}=4 y-8 ; & y(0)=0, & y(0)=-1, & y(0)=3\end{array}\) (b) \(\frac{d y}{d t}=y^{2}-4\) \(y(0)=-1, \quad y(0)=-3, \quad y(0)=4\) (c) \(\frac{d y}{d t}=(y-1)(y-2)(y+1) ; \quad y(0)=0, \quad y(0)=3\) \(\begin{array}{llll}\text { (d) } \frac{d y}{d t}=y^{2}+5 y-6 ; & y(0)=-5, & y(0)=-7, & y(0)=2\end{array}\)
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