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Chapter 8: Problem 9

Solve the differential equation. Be sure to check for possible constant solutions. If necessary, write your answer implicitly. $$ \frac{d y}{d t}=y+\frac{1}{y^{2}} $$

Short Answer

Expert verified
Constant solution is \( y = -1 \. Implicit solution is \left( \frac{y^3}{3} + y = \frac{y^3}{3} t + C \).

Step by step solution

01

Identify Constant Solutions

Find constant solutions by setting \(\frac{d y}{d t} = 0 \). This gives the equation \( y + \frac{1}{y^{2}} = 0 \). Solve for \( y \) to find any constant solutions.
02

Solve y + 1/y^2 = 0

Rearrange the equation \( y + \frac{1}{y^{2}} = 0 \) to \( y^3 + 1 = 0 \). Solve for \( y \), which gives the constant solution \( y = -1 \). Therefore, \( y = -1 \) is a constant solution.
03

Separate Variables

Rewrite the differential equation \( \frac{d y}{d t} = y + \frac{1}{y^2} \) as \( \frac{d y}{y + 1/y^2} = d t \). Separate variables: \( (y^2 + 1) d y = y^2 d t \).
04

Integrate Both Sides

Integrate both sides of the equation separately: \( \int (y^2 + 1) d y = \int y^2 d t \). This gives \( \frac{y^3}{3} + y = \frac{y^3}{3} t + C \), where \( C \) is the constant of integration.
05

Solve for y

The equation \( \frac{y^3}{3} + y = \frac{y^3}{3} t + C \) may be left in its implicit form, or it can be simplified further if necessary. Solve for \( y \) explicitly if possible.

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Key Concepts

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

Constant Solutions
A constant solution to a differential equation means that the solution does not change over time. For a differential equation of the form \(\frac{dy}{dt} = f(y)\), we find constant solutions by setting \(\frac{dy}{dt} = 0\). This turns the differential equation into an algebraic equation. In our example, the equation becomes \(y + \frac{1}{y^2} = 0\).
Solving \(y\textsuperscript{3} + 1 = 0\) gives \(y = -1\). Hence, \(y = -1\) is a constant solution. These solutions are important as they help set the groundwork for understanding the specific behaviors and patterns in differential equations.
Separation of Variables
The method of separation of variables allows us to rewrite a differential equation so that each variable appears on a different side of the equation. In our example, we start with the equation \(\frac{dy}{dt} = y + \frac{1}{y^2}\).
We can rearrange this to isolate \(dy\) and \(dt\) on opposite sides: \((y^2 + 1) \, dy = y^2 \, dt\). This makes integration possible for both sides independently. This method is helpful because it reduces the complexity of solving differential equations by dealing with each variable separately.
Integration
Once the variables are separated, the next step is to integrate both sides of the equation. For our differential equation, we integrate both sides like this: \(\int (y^2 + 1) \, dy = \int y^2 \, dt\).
This results in two integrals: \[ \frac{y^3}{3} + y = \frac{y^3}{3} t + C \] where \(C\) is the constant of integration that arises from indefinite integrals.
Integration helps us move from the differential form to the solution of the equation. Each side is independently solved for its integral, making it easier to find the relation between \(y\) and \(t\).
Implicit Solutions
An implicit solution to a differential equation is a form that includes both variables without isolating one. Sometimes, solving the differential equation explicitly for \(y\) isn't straightforward. An example from our differential equation is \[ \frac{y^3}{3} + y = \frac{y^3}{3} t + C \].
This form is called implicit because both \(t\) and \(y\) are present together, without solving for one in terms of the other.
Implicit solutions are often used when it's challenging to isolate the dependent variable. These types of solutions can still provide valuable insights into the behavior of the system described by the differential equation.

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Most popular questions from this chapter

Solve the initial-value problem. State an interval on which the solution exists. $$ t^{2} y^{\prime}+t y=t^{4} ; y(2)=5 $$

Let \(y(t)\) be the proportion of crystallizable fat in a sample after \(t\) hours. Then \(y\) satisfies the differential equation $$y^{\prime}=k\left(y^{n}-y\right).$$ where \(k\) is a constant and \(n\) is the Avrami exponent for decrystallization reactions. In practice, \(n\) is an integer greater than 1 that is computed from the time dependence of nucleation and the number of dimensions in which crystal growth occurs. \({ }^{15}\) Use this differential equation to solve Exercises \(29-32\). Although this differential equation for \(y\) is nonlinear, we will solve it using Theorem 3 after an initial modification. a) Divide both sides of the differential equation by \(y^{n} .\) Show that $$y^{-n} \frac{d y}{d t}+k y^{1-n}=k.$$ b) Let \(z(t)=[y(t)]^{1-n} .\) Use the Chain Rule to find \(z^{\prime}\) in terms of \(y\) and \(y^{\prime}\). c) Use the result of part (b) to show that $$\frac{d z}{d t}+k(1-n) z=k(1-n).$$ d) Find the general solution for \(z(t)\). e) Find the general solution for \(y(t)\). f) Suppose that the initial condition is \(y(0)=p\), where \(p\) is the initial proportion of crystallizable fat. Show that $$ y(t)=\left[1+\left(p^{1-n}-1\right) e^{(n-1) k t}\right]^{1 /(1-n)}. $$

Solve the differential equation. Be sure to check for possible constant solutions. If necessary, write your answer implicitly. $$ \frac{d y}{d t}=\frac{t}{\left(t^{2}+1\right)\left(y^{4}+1\right)} $$

Verify that the given function is a solution of the differential equation. $$ y^{\prime \prime}+4 y=4 \sin 2 x ; y=(1-x) \cos 2 x $$

Solve the higher-order initial-value problem. $$ f^{\prime \prime}(x)=\sin 3 x ; f(\pi)=-2 \text { and } f^{\prime}(\pi)=-3 $$
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