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Magazine Chapter 10: Problem 9
Solve the given equation using an integrating factor. Take \(t>0\). $$ y^{\prime}-2 t y=-4 t $$
Short Answer
Expert verified
The solution is \( y = 2 + Ce^{t^2} \)
Step by step solution
01
Identify the standard form
The given differential equation is \( y^{\backprime}-2 t y=-4 t \). The standard form for a first-order linear differential equation is \( y^{\backprime} + P(t)y = Q(t) \). Rewrite it as \( y^{\backprime} + (-2t)y = -4t \)
02
Determine the integrating factor
The integrating factor, \( \text{IF} \), is given by \( \text{IF} = e^{\backslashint P(t) \, dt} \). Here, \( P(t) = -2t \). Thus, the integrating factor is \[ e^{\backslashint -2t \, dt } = e^{-t^2} \]
03
Multiply through by the integrating factor
Multiply every term in the equation \( y^{\backprime} - 2ty = -4t \) by the integrating factor \( e^{-t^2} \) to obtain \[ e^{-t^2} y^{\backprime} - 2t e^{-t^2} y = -4t e^{-t^2} \]
04
Recognize the left-hand side as a derivative
The left-hand side of the equation \( e^{-t^2} y^{\backprime} - 2t e^{-t^2} y \) can be recognized as the derivative of the product \( ( e^{-t^2} y )^{\backprime} \). Thus, the equation becomes \[ ( e^{-t^2} y )^{\backprime} = -4t e^{-t^2} \]
05
Integrate both sides
Integrate both sides with respect to \( t \) to obtain \[ e^{-t^2} y = \backslashint -4t e^{-t^2} \, dt \]
06
Integrate the right-hand side
To integrate \( \backslashint -4t e^{-t^2} \, dt \), use the substitution \( u = t^2 \), hence \( du = 2t \, dt \. \) This gives \[ \backslashint -4t e^{-t^2} \, dt = \backslashint -2 e^{-u} \, du = 2 e^{-u} = 2 e^{-t^2} + C \]
07
Solve for \$ y \
Using the result from the integration, \[ e^{-t^2} y = 2 e^{-t^2} + C \] Divide through by \( e^{-t^2} \) to solve for \( y \), so \[ y = 2 + Ce^{t^2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
integrating factor
An integrating factor is a useful tool to solve linear differential equations. It simplifies the equation, making it easier to solve. For a first-order linear differential equation of the form \( y' + P(t)y = Q(t) \), the integrating factor, \( \text{IF} \), is calculated as follows:
\[ \text{IF} = e^{\int P(t) \, dt} \]
In our exercise, the equation is written as \( y' - 2ty = -4t \) and we identify \( P(t) = -2t \). Thus, to find the integrating factor:
\[ \text{IF} = e^{\int -2t \, dt} = e^{-t^2} \]
The integrating factor transforms the differential equation into an easier form to integrate.
\[ \text{IF} = e^{\int P(t) \, dt} \]
In our exercise, the equation is written as \( y' - 2ty = -4t \) and we identify \( P(t) = -2t \). Thus, to find the integrating factor:
\[ \text{IF} = e^{\int -2t \, dt} = e^{-t^2} \]
The integrating factor transforms the differential equation into an easier form to integrate.
differential equations
Differential equations are equations that involve an unknown function and its derivatives. They can be classified based on their order and linearity. A first-order differential equation involves only the first derivative of the function and the function itself.
In general, a first-order linear differential equation is of the form:
\[ y' + P(t)y = Q(t) \]
where \( y' \) denotes the derivative of \( y \) with respect to \( t \), and \( P(t) \) and \( Q(t) \) are given functions of \( t \). Solving such equations typically involves finding the integrating factor to simplify and integrate the equation.
In general, a first-order linear differential equation is of the form:
\[ y' + P(t)y = Q(t) \]
where \( y' \) denotes the derivative of \( y \) with respect to \( t \), and \( P(t) \) and \( Q(t) \) are given functions of \( t \). Solving such equations typically involves finding the integrating factor to simplify and integrate the equation.
integration by substitution
Integration by substitution is a technique used to solve more complicated integrals. It simplifies the integral by changing the variable of integration. In our example, we had to integrate \( \int -4t e^{-t^2} \, dt \).
To use substitution, set \( u = t^2 \), then \( du = 2t \, dt \). Therefore,
\[ \int -4t e^{-t^2} \, dt = \int -2 e^{-u} \, du \]
By changing the variables, the integral becomes simpler:
\[ \int -2 e^{-u} \, du = -2 \int e^{-u} \, du = -2 e^{-u} = -2 e^{-t^2} \]
Using substitution makes solving the integral more straightforward.
To use substitution, set \( u = t^2 \), then \( du = 2t \, dt \). Therefore,
\[ \int -4t e^{-t^2} \, dt = \int -2 e^{-u} \, du \]
By changing the variables, the integral becomes simpler:
\[ \int -2 e^{-u} \, du = -2 \int e^{-u} \, du = -2 e^{-u} = -2 e^{-t^2} \]
Using substitution makes solving the integral more straightforward.
solving linear ODEs
Solving linear ordinary differential equations (ODEs) involves several steps:
- The equation \( y' - 2ty = -4t \) was transformed to \( ( e^{-t^2}y )' = -4t e^{-t^2} \) using the integrating factor \( e^{-t^2} \).
- We then integrated both sides to get \( e^{-t^2}y = \int -4t e^{-t^2} \, dt \), which simplified using substitution.
- Finally, we solved for \( y \) and obtained \( y = 2 + Ce^{t^2} \), where \( C \) is an integration constant.
- Identify the standard form of the ODE: \( y' + P(t)y = Q(t) \).
- Find the integrating factor, \( \text{IF} = e^{\int P(t) \, dt} \).
- Multiply the entire differential equation by this integrating factor.
- Recognize the left-hand side as a derivative of a product.
- Integrate both sides of the equation.
- Solve for the unknown function \( y \).
- The equation \( y' - 2ty = -4t \) was transformed to \( ( e^{-t^2}y )' = -4t e^{-t^2} \) using the integrating factor \( e^{-t^2} \).
- We then integrated both sides to get \( e^{-t^2}y = \int -4t e^{-t^2} \, dt \), which simplified using substitution.
- Finally, we solved for \( y \) and obtained \( y = 2 + Ce^{t^2} \), where \( C \) is an integration constant.
