Problem 40 \(y^{(4)}-3 y^{\prime \prime \pr... [FREE SOLUTION]

Chapter 4: Problem 40

\(y^{(4)}-3 y^{\prime \prime \prime}+3 y^{\prime \prime}-y^{\prime}=6 t-20\)

Short Answer

Expert verified

The general solution to the differential equation is \(y(t) = c_1 + c_2t + c_2t^2 + c_2t^3 + 2t - 5\).

Step by step solution

Homogeneous Solution

First, find the characteristic equation of the homogeneous differential equation \(y^{(4)}-3 y^{\prime \prime \prime}+3 y^{\prime \prime}-y^{\prime}=0\) by replacing each derivative of \(y\) with \(r\) raised to the corresponding power and then solve for \(r\). We get \(r^4 - 3r^3 + 3r^2 - r = 0\) which factored gives \(r(r - 1)^3 = 0\). Thus, \(r=0\) and \(r=1\) (triple root). Based on these roots, the solution to the homogeneous differential equation is \(y_h(t) = c_1 + c_2t + c_2t^2 + c_2t^3\), where \(c_1, c_2, c_3, c_4\) are constants to be determined by initial conditions.

Particular Solution

For the non-homogeneous differential equation, try \(y_p(t) = At + B\) as a particular solution. Plug this into the differential equation to solve for \(A\) and \(B\). We get \(y_p^{(4)}-3 y_p^{\prime \prime \prime}+3 y_p^{\prime \prime}-y_p^{\prime}=6 t-20\). By simplifying we deduce \(A = 2\) and \(B = -5\). Therefore, the particular solution is \(y_p(t) = 2t - 5\).

General Solution

The general solution is a combination of the homogeneous solution and particular solution, \(y(t) = y_h(t) + y_p(t) = c_1 + c_2t + c_2t^2 + c_2t^3 + 2t - 5\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Homogeneous Differential Equation

A homogeneous differential equation is one where all terms are dependent on the function and its derivatives, with the absence of a free-standing constant or function on the right side. Effectively, it has the form like:

In our case: \( y^{(4)}-3y^{\prime \prime \prime} + 3y^{\prime \prime} - y^{\prime} = 0 \).
The important characteristic is that if \( y(t) \) is a solution, then any scaled version \( cy(t) \) (where \(c\) is a constant) is also a solution.

This creates a situation where the sum of multiple solutions is also a solution, reinforcing the term 'homogeneity'. Understanding this sets the stage for recognizing solution patterns that can simplify our work.

Characteristic Equation

The characteristic equation is crucial in solving homogeneous linear differential equations. It's formed by replacing derivatives in the equation with a variable raised to the corresponding power, such as \( r \). In our example, the differential equation transformed into:

\( r^4 - 3r^3 + 3r^2 - r = 0 \).
By factoring, we find the roots: \( r(r-1)^3 = 0 \), leading to roots \( r = 0 \) and \( r = 1 \) with multiplicity 3.

The roots tell us about the nature of the solutions. Specifically:

Each root corresponds to an exponential solution \( e^{rt} \), except this time it's the polynomial multiple due to the roots' multiplicity.

With a triple root \( r = 1 \), we derive solutions like \( c_2t + c_3t^2 + c_4t^3 \). This knowledge is essential for finding the homogeneous solution.

Particular Solution

The particular solution addresses any non-homogeneous components of the equation. The goal is to find a specific solution that satisfies the full differential equation, including its non-zero right-hand side.

In our case, the non-homogeneous equation is solved by proposing a particular solution form, such as \( y_p(t) = At + B \).
Substituting into the equation helps isolate terms to determine constants \( A \) and \( B \), resulting in \( A = 2 \) and \( B = -5 \), hence the particular solution of \( y_p(t) = 2t - 5 \).

Choosing the correct form requires recognizing patterns in the non-homogeneous term \( 6t - 20 \), which typically suggests polynomial solutions. The particular solution is necessary to solve the entire differential equation.

General Solution

The general solution combines both the homogeneous and particular solutions to provide the complete solution set of the differential equation. In simple terms, it sums everything up:

The homogeneous solution \( y_h(t) = c_1 + c_2t + c_2t^2 + c_2t^3 \) reflects all possible solutions from the characteristic equation.
The particular solution \( y_p(t) = 2t - 5 \) satisfies the additional terms from the non-homogeneous component.
The complete general solution is: \( y(t) = c_1 + c_2t + c_2t^2 + c_2t^3 + 2t - 5 \).

This approach allows for a flexible yet comprehensive solution, bridging the homogeneous and particular aspects of the equation. Constants \( c_1, c_2, c_3, \) etc., allow adaptations based on initial conditions.

91影视

\(y^{(4)}-3 y^{\prime \prime \prime}+3 y^{\prime \prime}-y^{\prime}=6 t-20\)

Short Answer

Step by step solution

Homogeneous Solution

Particular Solution

General Solution

Key Concepts

Homogeneous Differential Equation

Characteristic Equation

Particular Solution

General Solution

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