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Magazine Chapter 9: Problem 11
Solve the differential equations. \((t-1)^{3} \frac{d s}{d t}+4(t-1)^{2} s=t+1, \quad t>1\)
Short Answer
Expert verified
The solution is:
\[ s(t) = \frac{t^3/3 - t + C}{(t-1)^4}, \quad t > 1 \]
Step by step solution
01
Identify the type of differential equation
The given equation \[(t-1)^3 \frac{ds}{dt} + 4(t-1)^2 s = t + 1\] is a linear first-order differential equation. We can rewrite it in the standard form \[ \frac{ds}{dt} + P(t)s = Q(t) \] by dividing every term by \((t-1)^3\).
02
Divide the equation by \\(t-1\\)^3
Divide every term of the equation by \((t-1)^3\): \[ \frac{ds}{dt} + \frac{4}{(t-1)} s = \frac{t+1}{(t-1)^3} \] This equation is now in the standard form \[ \frac{ds}{dt} + P(t)s = Q(t) \] with \(P(t) = \frac{4}{(t-1)}\) and \(Q(t) = \frac{t+1}{(t-1)^3}\).
03
Find the integrating factor
The integrating factor \(\mu(t)\) is found using the formula \[ \mu(t) = e^{\int P(t) dt} \]. Calculate the integral: \[ \int \frac{4}{(t-1)} dt = 4 \ln |t-1| = \ln |(t-1)^4| \]. Therefore, the integrating factor is \[ \mu(t) = e^{\ln |(t-1)^4|} = (t-1)^4 \].
04
Multiply through by the integrating factor
Multiply every term in the differential equation by the integrating factor \((t-1)^4\): \[ (t-1)^4 \frac{ds}{dt} + 4(t-1)^3 s = \frac{(t+1)(t-1)}{1} \]. This simplifies to: \[ (t-1)^4 \frac{ds}{dt} + 4(t-1)^3 s = t^2 - 1^2 = t^2 - 1. \]
05
Integrate both sides
Notice that the left side of the equation is now the derivative of \((t-1)^4 s\). Therefore, we have: \[ \frac{d}{dt}[(t-1)^4 s] = t^2 - 1 \]. Integrate both sides: \[ (t-1)^4 s = \int (t^2 - 1) dt \]. Calculate the integral: \[ \int (t^2 - 1) dt = \frac{t^3}{3} - t + C \], where \(C\) is the constant of integration.
06
Solve for s(t)
Solve for \(s(t)\) by dividing both sides by \((t-1)^4\): \[ s(t) = \frac{t^3/3 - t + C}{(t-1)^4} \]. This provides the general solution for \(s(t)\) given \(t > 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Differential Equation
A first-order differential equation is a type of equation involving a function and its first derivative. Such equations are extremely important in various fields because they describe the rate of change in real-world applications, like population growth or cooling laws.
The equation in our exercise is a first-order differential equation due to the presence of the first derivative, \( \frac{ds}{dt} \). In general, the standard form for a first-order differential equation is \( \frac{dy}{dx} + P(x)y = Q(x) \).
Recognizing it's a first-order differential equation helps us know which solving techniques to apply, focusing on the rate of change as we see with the function \(s(t)\).
The equation in our exercise is a first-order differential equation due to the presence of the first derivative, \( \frac{ds}{dt} \). In general, the standard form for a first-order differential equation is \( \frac{dy}{dx} + P(x)y = Q(x) \).
Recognizing it's a first-order differential equation helps us know which solving techniques to apply, focusing on the rate of change as we see with the function \(s(t)\).
Integrating Factor
An integrating factor is a technique to solve linear first-order differential equations. It is a function that we multiply with the original equation to make its left-hand side easily integrable.
This method helps transform a product of a variable into a derivative of the product, simplifying the solving process.
An integrating factor, \(\mu(t)\), is found using the formula \( \mu(t) = e^{\int P(t) \, dt} \).
In our step-by-step solution, the integrating factor was calculated as \((t-1)^4\). This factor is pivotal as it allows us to rewrite the differential equation in an integrable form, making the left-hand side a perfect derivative of \((t-1)^4 s\).
This method helps transform a product of a variable into a derivative of the product, simplifying the solving process.
An integrating factor, \(\mu(t)\), is found using the formula \( \mu(t) = e^{\int P(t) \, dt} \).
In our step-by-step solution, the integrating factor was calculated as \((t-1)^4\). This factor is pivotal as it allows us to rewrite the differential equation in an integrable form, making the left-hand side a perfect derivative of \((t-1)^4 s\).
Linear Differential Equation
Linear differential equations are characterized by their linearity, meaning the unknown function and its derivatives appear to the power of one and don't get multiplied together.
In our example, the differential equation \((t-1)^3 \frac{ds}{dt} + 4(t-1)^2 s = t + 1\) is linear because the function \(s(t)\) and its derivative are both to the first power and not multiplied by each other.
Being linear means we can apply certain methods, like using an integrating factor, which is specific to linear forms. This simplicity eases the integration process, especially after transforming it with the integration factor, making it straightforward to integrate.
In our example, the differential equation \((t-1)^3 \frac{ds}{dt} + 4(t-1)^2 s = t + 1\) is linear because the function \(s(t)\) and its derivative are both to the first power and not multiplied by each other.
Being linear means we can apply certain methods, like using an integrating factor, which is specific to linear forms. This simplicity eases the integration process, especially after transforming it with the integration factor, making it straightforward to integrate.
Integration Technique
The integration technique involves integrating both sides of the equation after simplifying it. The goal is to find the solution function \( s(t) \).
In the given solution, integration is performed on the equation after transforming it using the integrating factor. This results in an expression that represents the antiderivative of the product \( (t-1)^4 s \).
Calculating the integral on the right side, as shown, gives \( \int (t^2 - 1) dt = \frac{t^3}{3} - t + C \).
In the given solution, integration is performed on the equation after transforming it using the integrating factor. This results in an expression that represents the antiderivative of the product \( (t-1)^4 s \).
Calculating the integral on the right side, as shown, gives \( \int (t^2 - 1) dt = \frac{t^3}{3} - t + C \).
- This is solved by having a constant of integration \( C \).
- Finally, to isolate \( s(t) \), we divide by \((t-1)^4\), yielding the general solution.
