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Magazine Chapter 10: Problem 8
Solve the given equation using an integrating factor. Take \(t>0\). $$y^{\prime}+2 t y=0$$
Short Answer
Expert verified
The solution is \( y = C e^{-t^2} \).
Step by step solution
01
Identify the Differential Equation
The given differential equation is: \[ y' + 2ty = 0 \]
02
Write the equation in standard form
Make sure the equation is in the form \( y' + p(t)y = q(t) \). Thus, we have: \[ y' + 2ty = 0 \]Here, \( p(t) = 2t \) and \( q(t) = 0 \).
03
Determine the Integrating Factor
The integrating factor (\( \mu(t) \)) is given by:\[ \mu(t) = e^{\int p(t) dt } \]For the given \( p(t) = 2t \):\[ \int 2t \, dt = t^2 \]Therefore,\[ \mu(t) = e^{t^2} \]
04
Multiply through by the Integrating Factor
Multiply the original differential equation by \( \mu(t) = e^{t^2} \):\[ e^{t^2} y' + e^{t^2} \, 2ty = 0 \]
05
Convert to an Exact Differential
Notice that the left-hand side of the equation is the derivative of \( y \, e^{t^2} \):\[ \frac{d}{dt}(y e^{t^2}) = 0 \]
06
Integrate Both Sides
Integrating both sides with respect to \( t \):\[ y e^{t^2} = \, C \]Where \( C \) is the constant of integration.
07
Solve for \( y \)
Solve for \( y \) by dividing both sides by \( e^{t^2} \):\[ y = \, C e^{-t^2} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
differential equation
A differential equation involves an unknown function and its derivatives. The equation we are working with is:
\[ y' + 2t y = 0 \]
Here, \(y'\) represents the first derivative of \(y\) with respect to \(t\). Differential equations can describe various physical phenomena such as motion, growth, and decay, making them essential in fields like physics, biology, and economics.
\[ y' + 2t y = 0 \]
Here, \(y'\) represents the first derivative of \(y\) with respect to \(t\). Differential equations can describe various physical phenomena such as motion, growth, and decay, making them essential in fields like physics, biology, and economics.
standard form
To solve a first-order linear differential equation using an integrating factor, it must be in the standard form:
\[ y' + p(t) y = q(t) \]
In our case,
\[ y' + 2t y = 0 \]
We can see that \(p(t) = 2t\) and \(q(t) = 0\). Getting the equation into this form is crucial for applying the integrating factor method efficiently.
\[ y' + p(t) y = q(t) \]
In our case,
\[ y' + 2t y = 0 \]
We can see that \(p(t) = 2t\) and \(q(t) = 0\). Getting the equation into this form is crucial for applying the integrating factor method efficiently.
integrating factor
An integrating factor is a function used to simplify integrating differential equations. It's defined for the equation
\[ y' + p(t)y = q(t) \]
as:
\[ \mu(t) = e^{\int p(t) dt} \]
For \(p(t) = 2t\),the integral is
\[ \int 2t \, dt = t^2 \]
Hence, the integrating factor is:
\[ \mu(t) = e^{t^2} \]
Multiplying the original differential equation by this integrating factor helps transform it into an exact differential, which we can then easily integrate.
\[ y' + p(t)y = q(t) \]
as:
\[ \mu(t) = e^{\int p(t) dt} \]
For \(p(t) = 2t\),the integral is
\[ \int 2t \, dt = t^2 \]
Hence, the integrating factor is:
\[ \mu(t) = e^{t^2} \]
Multiplying the original differential equation by this integrating factor helps transform it into an exact differential, which we can then easily integrate.
exact differential
Once we multiply the differential equation by the integrating factor, the left-hand side turns into an exact differential:
\[ e^{t^2} y' + e^{t^2} \cdot 2t y = 0 \]
Recognize and rewrite this as:
\[ \frac{d}{dt}(y e^{t^2}) = 0 \]
An exact differential implies that the left-hand side is the derivative of a product of functions鈥攊n this case, \(y e^{t^2}\). This is integral to solving the equation, as we can now easily integrate both sides.
\[ e^{t^2} y' + e^{t^2} \cdot 2t y = 0 \]
Recognize and rewrite this as:
\[ \frac{d}{dt}(y e^{t^2}) = 0 \]
An exact differential implies that the left-hand side is the derivative of a product of functions鈥攊n this case, \(y e^{t^2}\). This is integral to solving the equation, as we can now easily integrate both sides.
integration
To solve the differential equation, we integrate both sides with respect to \(t\):
\[ \int \frac{d}{dt}(y e^{t^2}) \, dt = \int 0 \, dt \]
This simplifies to:
\[ y e^{t^2} = C \]
where \(C\) is the constant of integration. To isolate \(y\), divide both sides by \(e^{t^2}\):
\[ y = C e^{-t^2} \]
This step completes the solution. Here, \(y\) is expressed explicitly in terms of \(t\), and \(C\) can be determined if an initial condition is provided.
\[ \int \frac{d}{dt}(y e^{t^2}) \, dt = \int 0 \, dt \]
This simplifies to:
\[ y e^{t^2} = C \]
where \(C\) is the constant of integration. To isolate \(y\), divide both sides by \(e^{t^2}\):
\[ y = C e^{-t^2} \]
This step completes the solution. Here, \(y\) is expressed explicitly in terms of \(t\), and \(C\) can be determined if an initial condition is provided.
