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Magazine Chapter 10: Problem 11
Solve the following differential equations: $$ y^{\prime}=3 t^{2} y^{2} $$
Short Answer
Expert verified
y = -\frac{1}{t^3 + D}
Step by step solution
01
Separate Variables
Rewrite the given differential equation to isolate variables on each side. Start with:y^{\text{prime}} = 3t^2 y^2y^{\text{prime}} is just \( \frac{dy}{dt} \), so we have:\( \frac{dy}{dt} = 3t^2 y^2 \).Separate the variables y and t:\( \frac{1}{y^2} dy = 3t^2 dt \)
02
Integrate Both Sides
Integrate both sides of the equation with respect to their respective variables:\(\int \frac{1}{y^2} dy = \int 3t^2 dt \).Which gives:\( \int y^{-2} dy = \int 3t^2 dt \)
03
Solve the Integrals
Integrate each side of the equation:\( \int y^{-2} dy = -y^{-1} + C_1 \) (where \( C_1 \) is the integration constant),and\( \int 3t^2 dt = t^3 + C_2 \) (where \( C_2 \) is the integration constant).
04
Combine the Results
Equate the results from the integral:\( -y^{-1} + C_1 = t^3 + C_2 \).For simplicity, combine the constants of integration into a single constant C. Set 1/C = D and rearrange the equation:\( -\frac{1}{y} = t^3 + D \)
05
Solve for y
Solve the equation for y to find the explicit solution:\( -\frac{1}{y} = t^3 + D \)Invert the function to solve for y:\( y = -\frac{1}{t^3 + D} \)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
separation of variables
Separation of variables is a powerful technique used to solve first-order differential equations. Here's how it works: Given a differential equation like \( \frac{dy}{dt} = 3t^2 y^2 \), we start by expressing it in a way that isolates the variables on opposite sides of the equation. In our case, we rewrite it as:
\( \frac{1}{y^2} dy = 3t^2 dt \).
The idea is to have all terms involving \( y \) on one side and all terms involving \( t \) on the other side. This method helps simplify solving the equation by integrating both sides separately.
\( \frac{1}{y^2} dy = 3t^2 dt \).
The idea is to have all terms involving \( y \) on one side and all terms involving \( t \) on the other side. This method helps simplify solving the equation by integrating both sides separately.
integration
Integration is the process of finding the integral of a function, which essentially is the reverse operation of differentiation. In our problem, once we separated the variables, the next step is to integrate both sides:
\( \frac{1}{y^2} dy = 3t^2 dt \).
We need to integrate both sides with respect to their respective variables:
\( \int \frac{1}{y^2} dy = \int 3t^2 dt \).
This gives us:
\( \int y^{-2} dy = -y^{-1} + C_1 \) and \( \int 3t^2 dt = t^3 + C_2 \).
The result of each integration step allows us to combine these results to find the general solution of the differential equation.
\( \frac{1}{y^2} dy = 3t^2 dt \).
We need to integrate both sides with respect to their respective variables:
\( \int \frac{1}{y^2} dy = \int 3t^2 dt \).
This gives us:
\( \int y^{-2} dy = -y^{-1} + C_1 \) and \( \int 3t^2 dt = t^3 + C_2 \).
The result of each integration step allows us to combine these results to find the general solution of the differential equation.
first-order differential equations
A first-order differential equation is an equation involving the first derivative of a function but no higher derivatives. The given equation (\( \frac{dy}{dt} = 3 t^2 y^2 \)) is a first-order differential equation because it involves only the first derivatives of y with respect to t.
Solving such equations typically involves the steps: separating variables, integrating both sides, and solving for the dependent variable.
Using separation of variables and then integrating enabled us to solve this first-order differential equation and obtain the solution.
Solving such equations typically involves the steps: separating variables, integrating both sides, and solving for the dependent variable.
Using separation of variables and then integrating enabled us to solve this first-order differential equation and obtain the solution.
integration constant
When integrating, you will often see a constant of integration added to the result. This is because integration determines the antiderivative of a function, which is not unique; it can differ by a constant. In our example, the integrations give us constants \(C_1\) and \(C_2\):
\( -y^{-1} + C_1 = t^3 + C_2 \).
These constants can be combined into a single constant like this:
\( -y^{-1} = t^3 + D \) (where \( D \) represents the combination of \( C_1 - C_2 \)).
This step is vital to forming a general solution representing all possible solutions to the differential equation.
\( -y^{-1} + C_1 = t^3 + C_2 \).
These constants can be combined into a single constant like this:
\( -y^{-1} = t^3 + D \) (where \( D \) represents the combination of \( C_1 - C_2 \)).
This step is vital to forming a general solution representing all possible solutions to the differential equation.
