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Magazine Chapter 10: Problem 20
Solve the given equation using an integrating factor. Take \(t>0\). $$\frac{1}{\sqrt{t+1}} y^{\prime}+y=1$$
Short Answer
Expert verified
The solution to the differential equation is \(y(t)=C \).
Step by step solution
01
Identify the Standard Form
Rewrite the differential equation in the form \(\frac{dy}{dt} + P(t)y = Q(t)\). We need to have the form \(\frac{dy}{dt} + P(t)y = Q(t)\). The given equation is \(\frac{1}{\frac{1}{\text{sqrt}(t+1)}} y' + y = 1\). Hence, identify \(P(t) = \frac{1}{\text{sqrt}(t+1)}\) and \(Q(t) = 1\).
02
Calculate the Integrating Factor
The integrating factor is given by the formula \(\text{IF} = e^{\text{int}(P(t)) \, dt}\). Here, \(P(t) =\frac{1}{\text{sqrt}(t+1)}\), so \(\text{IF} = e^{\text{int}\frac{1}{\text{sqrt}(t+1)} \, dt} = e^{2\text{sqrt}(t+1)} \).
03
Multiply Through by the Integrating Factor
Multiply both sides of the differential equation by the integrating factor \(e^{2\text{sqrt}(t+1)}\). The equation becomes: \(e^{2\text{sqrt}(t+1)} \frac{1}{\text{sqrt}(t+1)} y' + e^{2\text{sqrt}(t+1)}y = e^{2\text{sqrt}(t+1)}\).
04
Simplify and Integrate
Notice that the left side of the equation is the derivative of \(e^{2\text{sqrt}(t+1)} y\). So, the equation becomes: \(\frac{d}{dt} [e^{2\text{sqrt}(t+1)} y] = e^{2\text{sqrt}(t+1)}\). Integrate both sides with respect to \(t\): \(\text{int} [\frac{d}{dt} (e^{2\text{sqrt}(t+1)} y)] \, dt = \text{int} e^{2\text{sqrt}(t+1)} \, dt\). This implies \(e^{2\text{sqrt}(t+1)} y = \text{int} e^{2\text{sqrt}(t+1)} \, dt + C = \frac{\text{int} e^{2\text{sqrt}(t+1)}}{dt}\).
05
Final Solution
Integrate \(e^{2\text{sqrt}(t+1)}\) to find the antiderivative. Substitute the result back to get \(y(t)\). Hence, \(y(t) = \frac{e^{2\text{sqrt}(t+1)}}{2} + C \frac{e^{2\text{sqrt}(t+1)}}{2}\). So the general solution to the differential equation is \(y(t) = C e^{-2\text{sqrt}(t+1)} + \text{something}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Differential Equations
A differential equation relates a function with its derivatives. It expresses varying quantities using derivatives and is crucial for modeling real-world phenomena like growth rates and motion.
For example, an equation like \( y' + y = 1 \) connects the function \( y(t) \) with its derivative \( y' \).
This concept helps us predict future behavior of systems by solving these equations to find the unknown function \(y(t) \).
For example, an equation like \( y' + y = 1 \) connects the function \( y(t) \) with its derivative \( y' \).
This concept helps us predict future behavior of systems by solving these equations to find the unknown function \(y(t) \).
Integrating Factor
The integrating factor method is a vital technique to solve first-order linear differential equations of the form \( \frac{dy}{dt} + P(t) y = Q(t) \).
First, we find the integrating factor (IF), which transforms the differential equation into a form that's easier to integrate.
The integrating factor is given by:
\[ \text{IF} = e^{\int P(t) \, dt} \]
For our equation, \( P(t) = \frac{1}{\sqrt{t+1}} \) so the integrating factor is:
\[ \text{IF} = e^{\int \frac{1}{\sqrt{t+1}} \, dt} = e^{2 \sqrt{t+1}} \].
This helps simplify the left-hand side of our equation to a product rule form.
First, we find the integrating factor (IF), which transforms the differential equation into a form that's easier to integrate.
The integrating factor is given by:
\[ \text{IF} = e^{\int P(t) \, dt} \]
For our equation, \( P(t) = \frac{1}{\sqrt{t+1}} \) so the integrating factor is:
\[ \text{IF} = e^{\int \frac{1}{\sqrt{t+1}} \, dt} = e^{2 \sqrt{t+1}} \].
This helps simplify the left-hand side of our equation to a product rule form.
Antiderivatives
Antiderivatives, or indefinite integrals, reverse the process of differentiation. They are crucial when solving differential equations as they help find the original function from its derivative.
For instance, if we know the derivative \( f'(x) = e^{2\sqrt{x+1}} \), finding the antiderivative gives us:\[ \int e^{2\sqrt{x+1}} \, dx. \]
In our case, integrating \( e^{2\sqrt{t+1}} \) enables us to solve for \( y(t) \).
Antiderivatives include an arbitrary constant \(C\) because differentiation of a constant is zero, implying there can be infinitely many antiderivatives, differing only by a constant.
For instance, if we know the derivative \( f'(x) = e^{2\sqrt{x+1}} \), finding the antiderivative gives us:\[ \int e^{2\sqrt{x+1}} \, dx. \]
In our case, integrating \( e^{2\sqrt{t+1}} \) enables us to solve for \( y(t) \).
Antiderivatives include an arbitrary constant \(C\) because differentiation of a constant is zero, implying there can be infinitely many antiderivatives, differing only by a constant.
Initial Value Problems
Initial value problems (IVPs) are a type of differential equations where the solution must satisfy specific initial conditions. This makes the solution unique.
In our context, suppose we had an initial condition \( y(t_0) = y_0 \), this condition helps us find the particular solution by determining the constant \( C \).
By substituting the initial values into the general solution \( y(t) \, \) we solve for \( C \):
For example, if \( y(t) = C e^{-2 \sqrt{t+1}} + \frac{e^{2 \sqrt{t+1}}}{2} \) and \( y(t_0) = y_0 \), substituting these finds \( C \).
Once \( C \) is known, we have the unique solution catering to the given initial conditions.
In our context, suppose we had an initial condition \( y(t_0) = y_0 \), this condition helps us find the particular solution by determining the constant \( C \).
By substituting the initial values into the general solution \( y(t) \, \) we solve for \( C \):
For example, if \( y(t) = C e^{-2 \sqrt{t+1}} + \frac{e^{2 \sqrt{t+1}}}{2} \) and \( y(t_0) = y_0 \), substituting these finds \( C \).
Once \( C \) is known, we have the unique solution catering to the given initial conditions.
