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Chapter 2: Problem 20

Each \(D E\) in Problems \(15-22\) is a Bernoulli equation. In Problems 15-20, solve the given differential equation by using an appropriate substitution. $$ 3\left(1+t^{2}\right) \frac{d y}{d t}=2 t y\left(y^{3}-1\right) $$

Short Answer

Expert verified
Solve using substitution: let \( v = y^{-3} \) and follow through by solving the resulting linear equation.

Step by step solution

01

Identify the Bernoulli Equation

A Bernoulli differential equation has the form \( y' + P(x)y = Q(x)y^n \). The given equation is \( 3(1 + t^2) \frac{dy}{dt} = 2ty(y^3 - 1) \). Let's rewrite it as \( 3(1 + t^2) \frac{dy}{dt} = 2ty^4 - 2ty \). Dividing the whole equation by \(3(1 + t^2)\), we get \( \frac{dy}{dt} = \frac{2ty^4}{3(1 + t^2)} - \frac{2ty}{3(1 + t^2)} \), identifying \( n = 4 \).
02

Use Appropriate Substitution

To solve the Bernoulli equation, we use the substitution \( v = y^{1-n} = y^{-3} \). This gives us \( y = v^{-1/3} \), and the derivative \( \frac{dy}{dt} = -\frac{1}{3}v^{-4/3}\frac{dv}{dt} \). Substitute these into the rewritten equation.
03

Substitute and Simplify

Substitute \( y = v^{-1/3} \) and \( \frac{dy}{dt} = -\frac{1}{3}v^{-4/3}\frac{dv}{dt} \) into the differential equation: \( -\frac{1}{3}v^{-4/3}\frac{dv}{dt} = \frac{2t(v^{-4/3})}{3(1 + t^2)} - \frac{2t(v^{-1/3})}{3(1 + t^2)} \). Simplify this equation to find \( \frac{dv}{dt} \).
04

Solve the Linear Differential Equation

The resulting equation after substitution should be linear in terms of \( v \) and \( t \). Solve this linear differential equation for \( v \).
05

Substitute Back to Find y

After solving for \( v \), substitute back to get \( y \) using \( y = v^{-1/3} \). Simplify to find the solution in terms of \( y \) and \( t \).
06

Final Simplification and Result

Simplify the expression for \( y \) if needed to present the final solution. Ensure the solution satisfies the original differential equation.

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Key Concepts

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

Differential Equation
A differential equation is a mathematical equation involving derivatives, which represents how a function changes. In general, it looks like this: \( rac{dy}{dt} = f(t, y) \). This indicates how the variable \( y \) depends on another variable \( t \) and maybe \( y \) itself. Differential equations can be found everywhere in science and engineering because they express fundamental laws of nature, such as motion, heat flow, or population growth.The Bernoulli differential equation is a well-known type, characterized by having a non-linear term \( y^n \). It poses challenges that don't come with more straightforward linear equations, requiring specific methods to solve.
Substitution Method
Substitution is a technique that simplifies complex differential equations. When dealing with a Bernoulli differential equation of the form \( y' + P(x)y = Q(x)y^n \), the substitution \( v = y^{1-n} \) simplifies it into a linear form. The substitution method involves:
  • Altering the variable to make the equation easier to manage
  • Replacing the original equation with this new expression
This simplification leverages the properties of exponents to replace non-linear components with linear terms, making the equation solvable with standard methods.By using the substitution \( v = y^{-3} \), the Bernoulli equation is transformed into a linear differential equation, which is generally more straightforward to solve.
Solving Nonlinear Equations
Nonlinear equations, like Bernoulli's, contain terms like \( y^n \) that are not straightforward. They present a puzzle since typical linear techniques do not readily apply.To solve, a strategy is to transform the equation. The Bernoulli equation uses substitution to change the nature of the equation. Once the transformation is done:
  • The substitute variable \( v \) decreases the degree of non-linearity
  • It turns the equation into a linear form
Solve for the new variable. Finally, reverse the substitution to find the solution in terms of the original variable \( y \). This process demands careful manipulation of the mathematical expressions.
Advanced Mathematics
Advanced mathematics involves tackling complex problems using a wide array of techniques. Bernoulli equations are an example of this, showing how abstract principles are applied in problem-solving. In advanced mathematics:
  • We explore different forms of equations, like linear vs. nonlinear
  • Utilize substitution to simplify complex issues
  • Apply traditional methods to more sophisticated problems
The journey through solving a Bernoulli equation uses skills from both calculus and algebra, showing how these branches of mathematics come together to solve higher-level problems. It鈥檚 a perfect example of how mathematical theory becomes practical application.

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Most popular questions from this chapter

(a) The solution of the differential equation $$ \frac{2 x y}{\left(x^{2}+y^{2}\right)^{2}} d x+\left[1+\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right] d y=0 $$ is a family of curves that can be interpreted as streamlines of a fluid flow around a circular object whose boundary is described by the equation \(x^{2}+y^{2}=1\). Solve this \(\mathrm{DE}\) and note the solution \(f(x, y)=c\) for \(c=0\). (b) Use a CAS to plot the streamlines for \(c=0, \pm 0.2, \pm 0.4\) \(\pm 0.6\), and \(\pm 0.8\) in three different ways. First, use the contourplot of a CAS. Second, solve for \(x\) in terms of the variable \(y .\) Plot the resulting two functions of \(y\) for the given values of \(c\), and then combine the graphs. Third, use the CAS to solve a cubic equation for \(y\) in terms of \(x\).

The differential equation $$ \frac{d y}{d x}=\frac{-x+\sqrt{x^{2}+y^{2}}}{y} $$ describes the shape of a plane curve \(C\) that will reflect all incoming light beams to the same point and could be a model for the mirror of a reflecting telescope, a satellite antenna, or a solar collector. See Problem 29 in Exercises \(1.3\). There are several ways of solving this \(\mathrm{DE}\). (a) Verify that the differential equation is homogeneous (see Section 2.5). Show that the substitution \(y=u x\) yields $$ \frac{u d u}{\sqrt{1+u^{2}}\left(1-\sqrt{\left.1+u^{2}\right)}\right.}=\frac{d x}{x}. $$ Use a CAS (or another judicious substitution) to integrate the left-hand side of the equation. Show that the curve \(C\) must be a parabola with focus at the origin and is symmetric with respect to the \(x\) -axis. (b) Show that the first differential equation can also be solved by means of the substitution \(u=x^{2}+y^{2}\).

Suppose an \(R C\) -series circuit has a variable resistor. If the resistance at time \(t\) is given by \(R=k_{1}+k_{2} t\), where \(k_{1}\) and \(k_{2}\) are known positive constants, then (9) becomes $$ \left(k_{1}+k_{2} t\right) \frac{d q}{d t}+\frac{1}{C} q=E(t) $$ If \(E(t)=E_{0}\) and \(q(0)=q_{0}\), where \(E_{0}\) and \(q_{0}\) are constants, show that $$ q(t)=E_{0} C+\left(q_{0}-E_{0} C\right)\left(\frac{k_{1}}{k_{1}+k_{2} t}\right)^{1 / C k_{2}} $$

Solve the given initial-value problem. Give the largest interval \(I\) over which the solution is defined. $$ L \frac{d i}{d t}+R i=E ; \quad i(0)=i_{0}, L, R, E, \text { and } i_{0} \text { constants } $$

When forgetfulness is taken into account, the rate of memorization of a subject is given by $$ \frac{d A}{d t}=k_{1}(M-A)-k_{2} A $$ where \(k_{1}>0, k_{2}>0, A(t)\) is the amount to be memorized in time \(t, M\) is the total amount to be memorized, and \(M-A\) is the amount remaining to be memorized. See Problems 25 and 26 in Exercises \(1.3\). (a) Since the DE is autonomous, use the phase portrait concept of Section \(2.1\) to find the limiting value of \(A(t)\) as \(t \rightarrow \infty\). Interpret the result. (b) Solve for \(A(t)\) subject to \(A(0)=0\). Sketch the graph of \(A(t)\) and verify your prediction in part (a).
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