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Chapter 10: Problem 4

Solve the following differential equations: $$ \frac{d y}{d t}=-\frac{1}{t^{2} y^{2}} $$

Short Answer

Expert verified
The solution to the differential equation is \( y = \sqrt[3]{3 \left( -\frac{1}{t} + C \right)} \).

Step by step solution

01

Separate the variables

Reorganize the equation so that all terms involving y are on one side and all terms involving t are on the other side. Start with the given differential equation: \[\frac{dy}{dt} = -\frac{1}{t^2 y^2}\] Multiply both sides by \(y^2\) and \(dt\): \[y^2 \, dy = -\frac{1}{t^2} \, dt\]
02

Integrate both sides

Integrate both sides of the equation to solve for y as a function of t. \[\int y^2 \, dy = \int -\frac{1}{t^2} \, dt\] Perform the integration: \[\frac{y^3}{3} = \frac{1}{t} + C\] where C is the constant of integration.
03

Solve for y

Multiply both sides by 3 to isolate \(y^3\): \[y^3 = 3 \left(-\frac{1}{t} + C\right)\] Take the cube root of both sides to solve for y: \[y = \sqrt[3]{3 \left( -\frac{1}{t} + C \right)}\]

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Key Concepts

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

separable differential equations
A separable differential equation is a type of equation that can be rewritten to separate the variables on different sides. This makes it more manageable to solve. In our example, the differential equation is given as: \(\frac{dy}{dt} = -\frac{1}{t^2 y^2}\).To separate the variables, we multiply both sides by \(y^2\) and \(dt\). This step gives us:\(y^2 \, dy = -\frac{1}{t^2} \, dt\).
Now, all terms with \(y\) are on one side, and all terms with \(t\) are on the other. This allows us to integrate each side individually, which is the next step.
integration
Integration is a fundamental concept in calculus used to find the antiderivative or the area under a curve. After separating the variables in our differential equation, we integrate both sides. We start with:\(\int y^2 \, dy = \int -\frac{1}{t^2} \, dt\).
To perform the integration, remember the basic rules:
  • The antiderivative of \(y^2\) is \(\frac{y^3}{3}\).
  • The antiderivative of \(-\frac{1}{t^2}\) is \(\frac{1}{t}\).
The integration results in:\( \frac{y^3}{3} = \frac{1}{t} + C \), where \(C\) is our constant of integration.
constant of integration
When integrating, we add a constant of integration, \(C\), because the integral of a function is not unique. The derivative of a constant is zero, meaning when taking the antiderivative, we might have missed a constant that was in the original function.
After integrating both sides of our separated equation, we get:\( \frac{y^3}{3} = \frac{1}{t} + C \).
Here, \(C\) represents an unknown constant that we determine by using initial conditions or additional information about the solution.
solving for a variable
The final step is solving for the variable \(y\). Our integrated equation is:\( \frac{y^3}{3} = \frac{1}{t} + C \).
First, we multiply both sides by 3 to isolate \(y^3\):\( y^3 = 3 \, ( -\frac{1}{t} + C )\).
Then, we take the cube root of both sides to solve for \(y\):\( y = \, \root 3 \, {3 \, ( -\frac{1}{t} + C )} \).
Now, we've solved the differential equation for \(y\) in terms of \(t\), providing a complete solution. Each step makes the problem more manageable by breaking it down into simpler parts.

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Most popular questions from this chapter

One or more initial conditions are given for each differential equation in the following exercises. Use the qualitative theory of autonomous differential equations to sketch the graphs of the corresponding solutions. Include a \(y z\) -graph if one is not already provided. Always indicate the constant solutions on the \(t y\) -graph whether they are mentioned or not. $$ y^{\prime}=y^{3}, y(0)=-1, y(0)=1 $$

A body was found in a room when the room's temperature was \(70^{\circ} \mathrm{F}\). Let \(f(t)\) denote the temperature of the body \(t\) hours from the time of death. According to Newton's law of cooling, \(f\) satisfies a differential equation of the form $$ y^{\prime}=k(T-y) $$ (a) Find \(T\). (b) After several measurements of the body's temperature, it was determined that when the temperature of the body was 80 degrees it was decreasing at the rate of 5 degrees per hour. Find \(k\). (c) Suppose that at the time of death the body's temperature was about normal, say \(98^{\circ} \mathrm{F}\). Determine \(f(t)\). (d) When the body was discovered, its temperature was \(85^{\circ} \mathrm{F}\). Determine how long ago the person died.

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}=e^{-t}(y+1) $$

Solve the initial-value problem. $$ t y^{\prime}+y=\sin t, y\left(\frac{\pi}{2}\right)=0, t>0 $$

Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}-2 t y=-4 t $$
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