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Magazine Chapter 2: Problem 28
\(d y / d t=-y^{3}, y(0)=a\)
Short Answer
Expert verified
y(t) = \pm \sqrt{\frac{1}{\frac{1}{a^2} + 2t}}
Step by step solution
01
- Identify the Type of Differential Equation
This is a separable differential equation because it can be written in the form of \[ \frac{dy}{dt} = -y^3 \]
02
- Separate Variables
Rewrite the equation by separating the variables y and t on different sides:\[ \frac{dy}{-y^3} = dt \]
03
- Integrate Both Sides
Integrate both sides of the equation:\[\int \frac{dy}{-y^3} = \int dt\]
04
- Simplify the Integrals
Simplify and solve the integral:\[\int -y^{-3} dy = \int dt\] The left side becomes:\[\int -y^{-3} dy = \int -y^{-3} dy = \frac{y^{-2}}{2} = -\frac{1}{2y^2}\]So the equation becomes:\[-\frac{1}{2y^2} = t + C\]where C is the constant of integration.
05
- Solve for the Constant of Integration
Use the given initial condition y(0) = a to solve for C:\[-\frac{1}{2a^2} = 0 + C\]Thus, C = -\frac{1}{2a^2}.
06
- Solve for y(t)
Substitute C back into the equation:\[-\frac{1}{2y^2} = t - \frac{1}{2a^2}\]Now, solve for y:\[\frac{1}{2y^2} = \frac{1}{2a^2} + t\]Invert and simplify:\[y^2 = \frac{1}{\frac{1}{a^2} + 2t}\]Finally, take the square root:\[y(t) = \pm \sqrt{\frac{1}{\frac{1}{a^2} + 2t}}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Initial Value Problem
An initial value problem specifies a condition that the solution to a differential equation must satisfy at a particular point. In this case, we have:
- Differential equation: \(\frac{dy}{dt} = -y^3\)
- Initial value: \(y(0) = a\)
Integration
Integration is the process of finding an integral, which is the inverse operation of differentiation. When solving a separable differential equation, we integrate both sides after separating the variables:
\[ \begin{align*} \text{Separate variables:} & \frac{dy}{-y^3} = dt \ \text{Integrate both sides:} & \int \frac{dy}{-y^3} = \int dt \ \text{Simplify the integrals:} & \int -y^{-3} dy = \int dt \end{align*}\]
The left side becomes:
\[ \begin{align*} \int -y^{-3} dy &= \frac{y^{-2}}{2} \- \frac{1}{2y^2} \end{align*} \]
And for the right side, integrating \(dt\) simply adds a constant of integration \(C\):
\[ -\frac{1}{2y^2} = t + C \]
\[ \begin{align*} \text{Separate variables:} & \frac{dy}{-y^3} = dt \ \text{Integrate both sides:} & \int \frac{dy}{-y^3} = \int dt \ \text{Simplify the integrals:} & \int -y^{-3} dy = \int dt \end{align*}\]
The left side becomes:
\[ \begin{align*} \int -y^{-3} dy &= \frac{y^{-2}}{2} \- \frac{1}{2y^2} \end{align*} \]
And for the right side, integrating \(dt\) simply adds a constant of integration \(C\):
\[ -\frac{1}{2y^2} = t + C \]
Variable Separation
Variable separation is a technique used to solve differential equations by rearranging terms so that each side contains only one variable. In our example:
We start with the differential equation: \( \frac{dy}{dt} = -y^3 \)
Separate the variables by moving \(y\) terms to one side and \(t\) terms to the other side:
\[ \frac{dy}{-y^3} = dt \]
This equation now allows us to integrate both sides independently, which leads to the next step.
We start with the differential equation: \( \frac{dy}{dt} = -y^3 \)
Separate the variables by moving \(y\) terms to one side and \(t\) terms to the other side:
\[ \frac{dy}{-y^3} = dt \]
This equation now allows us to integrate both sides independently, which leads to the next step.
Constant of Integration
When we integrate a function, we must add a constant of integration, \(C\), because the antiderivative of a function is not unique. It represents an entire family of functions that differ by a constant.
During the integration:
\[ -\frac{1}{2y^2} = t + C \]
We use the initial condition \( y(0) = a \) to solve for \( C \):
\[ -\frac{1}{2a^2} = 0 + C \]
So, \( C = -\frac{1}{2a^2} \).
Substituting \( C \) back into our equation leads us to the particular solution that satisfies the initial condition.
During the integration:
\[ -\frac{1}{2y^2} = t + C \]
We use the initial condition \( y(0) = a \) to solve for \( C \):
\[ -\frac{1}{2a^2} = 0 + C \]
So, \( C = -\frac{1}{2a^2} \).
Substituting \( C \) back into our equation leads us to the particular solution that satisfies the initial condition.
