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Magazine Chapter 4: Problem 3
\(y^{(5)}-y^{(4)}=1\)
Short Answer
Expert verified
The general solution is \( y = -\frac{x^4}{24} + Ce^{x} + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \).
Step by step solution
01
Identify the given differential equation
Consider the differential equation given: \(y^{(5)} - y^{(4)} = 1\)
02
Introduce a new variable
Let \(z = y^{(4)}\). This substitution transforms the original fifth-order differential equation into a first-order differential equation involving \(z\). Thus, the equation becomes: \( z' - z = 1 \)
03
Solve the first-order differential equation
To solve the differential equation \( z' - z = 1 \), recognize that it is a linear first-order differential equation of the form \(z' + P(z) = Q\), where \(P = -1\) and \(Q = 1\). The integrating factor (IF) is: \[ IF = e^{\int -1 \, dx} = e^{-x} \] Multiply through by the integrating factor: \[ e^{-x} z' - e^{-x} z = e^{-x} \] Simplifying, we get: \[ \frac{d}{dx}(e^{-x} z) = e^{-x} \]
04
Integrate both sides to find z
Integrate both sides with respect to \(x\): \[ \int \frac{d}{dx}(e^{-x} z) \, dx = \int e^{-x} \, dx \] Using the Fundamental Theorem of Calculus, we get: \[ e^{-x} z = -e^{-x} + C \] Therefore, \( z = -1 + Ce^{x} \)
05
Substitute back for y
Recall that \(z = y^{(4)}\), so we have: \( y^{(4)} = -1 + Ce^{x} \)
06
Integrate to find y
Integrate repeatedly to find \(y\): \[ y^{(3)} = \int (-1 + Ce^{x}) \, dx = -x + Ce^{x} + C_1 \] Integrate again: \[ y'' = \int (-x + Ce^{x} + C_1) \, dx = -\frac{x^2}{2} + Ce^{x} + C_1 x + C_2 \] Integrate again: \[ y' = \int (-\frac{x^2}{2} + Ce^{x} + C_1 x + C_2) \, dx = -\frac{x^3}{6} + Ce^{x} + \frac{C_1 x^2}{2} + C_2 x + C_3 \] Integrate one last time: \[ y = \int (-\frac{x^3}{6} + Ce^{x} + \frac{C_1 x^2}{2} + C_2 x + C_3) \, dx = -\frac{x^4}{24} + Ce^{x} + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Linear First-Order Differential Equation
In the given problem, we start with the fifth-order differential equation:
\(y^{(5)} - y^{(4)} = 1\).
To simplify, we introduce a new variable, \(z = y^{(4)}\).
This turns our complex equation into a first-order differential equation:
\(z' - z = 1\).
Linear first-order differential equations are of the form:
\(z' + P(z) = Q\),
where \(P\) and \(Q\) are functions of \(x\).
In our example:
\(y^{(5)} - y^{(4)} = 1\).
To simplify, we introduce a new variable, \(z = y^{(4)}\).
This turns our complex equation into a first-order differential equation:
\(z' - z = 1\).
Linear first-order differential equations are of the form:
\(z' + P(z) = Q\),
where \(P\) and \(Q\) are functions of \(x\).
In our example:
- \(P = -1\)
- \(Q = 1\)
Integrating Factor
An integrating factor (IF) is a function that simplifies the process of solving linear first-order differential equations.
For the general form \(z' + P(z) = Q\), the integrating factor is calculated as:
\[ IF = e^{\int P \, dx} \]
In our specific case, \(P = -1\), so:
\[ IF = e^{\int -1 \, dx} = e^{-x} \]
We multiply the original equation \(z' - z = 1\) by this factor:
\[ e^{-x} z' - e^{-x} z = e^{-x} \]
which simplifies to:
\[ \frac{d}{dx}(e^{-x} z) = e^{-x} \]
By doing so, the differential equation is transformed into a simple one that can be integrated straightforwardly.
For the general form \(z' + P(z) = Q\), the integrating factor is calculated as:
\[ IF = e^{\int P \, dx} \]
In our specific case, \(P = -1\), so:
\[ IF = e^{\int -1 \, dx} = e^{-x} \]
We multiply the original equation \(z' - z = 1\) by this factor:
\[ e^{-x} z' - e^{-x} z = e^{-x} \]
which simplifies to:
\[ \frac{d}{dx}(e^{-x} z) = e^{-x} \]
By doing so, the differential equation is transformed into a simple one that can be integrated straightforwardly.
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus bridges the gap between differentiation and integration.
SPECIFICALLY, IT STATES TWO MAIN THINGS:
1. Anti-differentiation (or indefinite integral) undoes differentiation.
2. The accumulation of 'area under the curve' can be determined by evaluating the anti-derivative at the boundaries.
In our case, integrating both sides:
\[ \int \frac{d}{dx}(e^{-x} z) \, dx = \int e^{-x} \, dx \]
we get:
\[ e^{-x} z = -e^{-x} + C \]
where \(C\) is the constant of integration.
Thus, solving for \(z\), we find:
\[ z = -1 + Ce^{x} \].
SPECIFICALLY, IT STATES TWO MAIN THINGS:
1. Anti-differentiation (or indefinite integral) undoes differentiation.
2. The accumulation of 'area under the curve' can be determined by evaluating the anti-derivative at the boundaries.
In our case, integrating both sides:
\[ \int \frac{d}{dx}(e^{-x} z) \, dx = \int e^{-x} \, dx \]
we get:
\[ e^{-x} z = -e^{-x} + C \]
where \(C\) is the constant of integration.
Thus, solving for \(z\), we find:
\[ z = -1 + Ce^{x} \].
Repeated Integration
To solve for \(y\), we need repeated integration starting from \(y^{(4)} = z \).
We have:
\( z = -1 + Ce^{x} \)
Integrate once to find \( y^{(3)} \):
\[ y^{(3)} = \int (-1 + Ce^{x}) \, dx = -x + Ce^{x} + C_1 \]
Integrate again for \( y'' \):
\[ y'' = \int (-x + Ce^{x} + C_1) \, dx = -\frac{x^2}{2} + Ce^{x} + C_1 x + C_2 \]
Then integrate for \( y' \):
\[ y' = \int (-\frac{x^2}{2} + Ce^{x} + C_1 x + C_2) \, dx = -\frac{x^3}{6} + Ce^{x} + \frac{C_1 x^2}{2} + C_2 x + C_3 \]
Finally, integrate for \( y \):
\[ y = \int (-\frac{x^3}{6} + Ce^{x} + \frac{C_1 x^2}{2} + C_2 x + C_3) \, dx = -\frac{x^4}{24} + Ce^{x} + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \]
You now have the general solution for \(y\)!
We have:
\( z = -1 + Ce^{x} \)
Integrate once to find \( y^{(3)} \):
\[ y^{(3)} = \int (-1 + Ce^{x}) \, dx = -x + Ce^{x} + C_1 \]
Integrate again for \( y'' \):
\[ y'' = \int (-x + Ce^{x} + C_1) \, dx = -\frac{x^2}{2} + Ce^{x} + C_1 x + C_2 \]
Then integrate for \( y' \):
\[ y' = \int (-\frac{x^2}{2} + Ce^{x} + C_1 x + C_2) \, dx = -\frac{x^3}{6} + Ce^{x} + \frac{C_1 x^2}{2} + C_2 x + C_3 \]
Finally, integrate for \( y \):
\[ y = \int (-\frac{x^3}{6} + Ce^{x} + \frac{C_1 x^2}{2} + C_2 x + C_3) \, dx = -\frac{x^4}{24} + Ce^{x} + \frac{C_1 x^3}{6} + \frac{C_2 x^2}{2} + C_3 x + C_4 \]
You now have the general solution for \(y\)!
