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Magazine Chapter 16: Problem 12
Solve the differential equations $$(t+1) \frac{d s}{d t}+2 s=3(t+1)+\frac{1}{(t+1)^{2}}, \quad t>-1$$
Short Answer
Expert verified
The general solution is \(s(t) = \frac{3}{4}(t+1)^2 + \frac{t}{(t+1)^2} + \frac{C}{(t+1)^2}\).
Step by step solution
01
Identify type of differential equation
The given differential equation is \[(t+1) \frac{d s}{d t} + 2s = 3(t+1) + \frac{1}{(t+1)^{2}}\]This is a first-order linear differential equation in the standard form \[a(t) \frac{ds}{dt} + b(t)s = c(t)\] where \[a(t) = t+1, \ b(t) = 2, \ c(t) = 3(t+1) + \frac{1}{(t+1)^2}\]We need to solve this equation by finding an integrating factor.
02
Compute the integrating factor
To find the integrating factor \(\mu(t)\), we use the formula:\[\mu(t) = e^{\int \frac{b(t)}{a(t)} dt}\] with \(b(t) = 2\) and \(a(t) = t+1\).Calculate the integral:\[\frac{b(t)}{a(t)} = \frac{2}{t+1}\]\[\int \frac{2}{t+1} dt = 2\ln|t+1| = \ln|(t+1)^2|\]\[\mu(t) = e^{\ln|(t+1)^2|} = (t+1)^2\]Thus, the integrating factor \(\mu(t) = (t+1)^2\).
03
Multiply the differential equation by the integrating factor
Multiply the entire differential equation by \((t+1)^2\) to facilitate integration.\[(t+1)^3 \frac{ds}{dt} + 2(t+1)^2s = 3(t+1)^3 + 1\]
04
Transform to exact equation and integrate
The transformed equation can be rewritten as:\[\frac{d}{dt}[(t+1)^2s] = 3(t+1)^3 + 1\]Integrate both sides with respect to \(t\):\[\int \frac{d}{dt}[(t+1)^2s] dt = \int (3(t+1)^3 + 1) dt\]On the left side, integration gives us:\[(t+1)^2s\]On the right side:\[\int 3(t+1)^3 dt + \int 1 dt = \left(\frac{3}{4}(t+1)^4\right) + t\]
05
Solve for s(t) and apply integration constant
Thus:\[(t+1)^2s = \frac{3}{4}(t+1)^4 + t + C\]Solve for \(s(t)\) by dividing both sides by \((t+1)^2\):\[s(t) = \frac{3}{4}(t+1)^2 + \frac{t}{(t+1)^2} + \frac{C}{(t+1)^2}\]
06
Solution verification
The general solution is given by:\[s(t) = \frac{3}{4}(t+1)^2 + \frac{t}{(t+1)^2} + \frac{C}{(t+1)^2}\]This satisfies the original differential equation for any constant \(C\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
First-Order Linear Differential Equation
A first-order linear differential equation is a type of equation that involves the first derivative of the unknown function. This is solved relatively easily compared to higher-order equations. These equations are expressed in the form:
In our exercise, we identified this with \(a(t) = t+1, b(t) = 2, \) and \(c(t) = 3(t+1) + \frac{1}{(t+1)^2} \). Understanding the standard form helps in applying methods for solving, like using an integrating factor, which we'll explore next.
- \( a(t) \frac{ds}{dt} + b(t)s = c(t) \)
In our exercise, we identified this with \(a(t) = t+1, b(t) = 2, \) and \(c(t) = 3(t+1) + \frac{1}{(t+1)^2} \). Understanding the standard form helps in applying methods for solving, like using an integrating factor, which we'll explore next.
Integrating Factor
The integrating factor is a powerful tool in solving first-order linear differential equations. It transforms the equation into a form where integration becomes straightforward. You compute the integrating factor, denoted \(\mu(t)\), using:
In the example provided, \(b(t) = 2\) and \(a(t) = t+1\). The process involved calculating the integral:
- \( \mu(t) = e^{\int \frac{b(t)}{a(t)} dt} \)
In the example provided, \(b(t) = 2\) and \(a(t) = t+1\). The process involved calculating the integral:
- \( \int \frac{2}{t+1} dt = 2\ln|t+1| \)
- \( \mu(t) = (t+1)^2 \)
General Solution
Finding the general solution requires integrating the transformed differential equation and solving for the unknown function. After applying the integrating factor, the equation becomes an exact form:
- \( \frac{d}{dt}[(t+1)^2s] = 3(t+1)^3 + 1 \)
- Left side: \( (t+1)^2s \)
- Right side: \( \frac{3}{4}(t+1)^4 + t + C \)
- \( s(t) = \frac{3}{4}(t+1)^2 + \frac{t}{(t+1)^2} + \frac{C}{(t+1)^2} \)
