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Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}-8 y^{\prime}=0 $$

Short Answer

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Question: Determine the general solution of the given differential equation: $$ y^{\mathrm{iv}}-8 y^{\prime}=0 $$ Answer: The general solution of the given differential equation is $$ y(x) = C_1 + C_2 e^{2x} + C_3 x e^{2x} + C_4 x^2 e^{2x}. $$

Step by step solution

01

Identify the differential equation type

The given differential equation is a homogeneous linear ordinary differential equation of order 4: $$ y^{\mathrm{iv}}-8 y^{\prime}=0 $$
02

Create the characteristic equation

To create the characteristic equation, we need to replace the derivatives with exponentials. For the given equation, we will get: $$ r^4 - 8r = 0 $$
03

Solve the characteristic equation

We'll now solve the characteristic equation: $$ r^4 - 8r = 0 $$ Factor out r, and we get: $$ r(r^3 - 8) = 0 $$ The first root is clearly \(r_1 = 0\). For the cubic equation \(r^3 - 8 = 0\), rearrange to get: $$ r^3 = 8 $$ Now we find the other three roots: $$ r_{2,3,4} = \sqrt[3]{8} = 2 $$ So, the characteristic roots are \(r_1 = 0\) and \(r_{2,3,4} = 2\)
04

Construct the general solution

Based on the roots we found in step 3, we can now write the general solution: $$ y(x) = C_1 + C_2 e^{2x} + C_3 x e^{2x} + C_4 x^2 e^{2x} $$ This is the general solution of the given fourth-order homogeneous linear ordinary differential equation.

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Key Concepts

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

Characteristic Equation
When faced with a homogeneous linear ordinary differential equation, one of the pivotal steps toward solving it involves formulating what is known as the characteristic equation. This equation translates the differential equation into algebraic terms, leveraging the power of exponentials to simplify the process of finding solutions. In the context of our example, the fourth order differential equation, \( y^{\mathrm{iv}}-8 y^{\prime}=0 \), is transformed by equating the terms involving derivatives to a polynomial with a variable typically denoted as \(r\). This yields the equation \( r^4 - 8r = 0 \).

By solving this polynomial equation, you establish the roots, which correspond to the exponentials in the homogeneous solution to the differential equation. In essence, the characteristic equation acts as a bridge that takes us from a complex differential equation to a more manageable algebraic one, rendering the path to the solution more accessible.
General Solution of Differential Equation
The general solution of a differential equation is a formula that incorporates all possible solutions to the equation. It embodies the complete set of solutions, with the inclusion of arbitrary constants corresponding to the initial conditions of the problem, which are named C1, C2, and so on. This variety allows the general solution to be tailored for particular scenarios, where initial values are specified.

For the differential equation in question, once the characteristic equation is solved and the roots are determined, you craft the general solution. This involves constructing a linear combination of functions that are exponential in nature, each raised to the power of a found root, multiplied by these arbitrary constants. In our step by step example, the general solution is expressed as \( y(x) = C_1 + C_2 e^{2x} + C_3 x e^{2x} + C_4 x^2 e^{2x} \), which ultimately encapsulates all possible behaviors of the system represented by the original differential equation.
Fourth-Order Differential Equation
Differential equations vary in complexity, with the order denoting the highest derivative present. A fourth-order differential equation involves the fourth derivative of the unknown function, making it significantly more complicated than its lower-order counterparts. These equations often arise in contexts like beam deflection in engineering and the motion of a vibrating string, embodying complex dynamic systems.

To solve these higher-order equations, we apply methods generalizable across differing orders, starting with the formulation of the characteristic equation, as previously discussed. After determining the roots, the general solution is constructed with respect to these roots. Despite the higher complexity, the structured approach to solving ensures that fourth-order differential equations, such as \( y^{\mathrm{iv}}-8 y^{\prime}=0 \), become tractable, allowing for detailed analysis of intricate physical phenomena.

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Most popular questions from this chapter

The purpose of this problem is to show that if \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0\) for some \(t_{0}\) in an interval \(I,\) then \(y_{1}, \ldots, y_{n}\) are linearly independent on \(I,\) and if they are linearly independent and solutions of $$ L(y]=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y=0 $$ on \(I,\) then \(W\left(y_{1}, \ldots, y_{n}\right)\) is nowhere zero in \(I .\) (a) Suppose that \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right) \neq 0,\) and suppose that $$ c_{1} y_{1}(t)+\cdots+c_{n} y_{n}(t)=0 $$ for all \(t\) in \(I\). By writing the equations corresponding to the first \(n-1\) derivatives of Fa. (ii) at \(t_{0}\), show that \(c_{1}=\cdots=c_{n}=0 .\) Therefore, \(y_{1}, \ldots, y_{n}\) are linearly independent. (b) Suppose that \(y_{1}, \ldots, y_{n}\) are linearly independent solutions of Eq. (i). If \(W\left(y_{1}, \ldots, y_{n}\right)\left(t_{0}\right)=0\) for some \(t_{0},\) show that there is a nonzero solution of Eq. (i) satisfying the initial conditions $$ y\left(t_{0}\right)=y^{\prime}\left(t_{0}\right)=\cdots=y^{(n-1)}\left(t_{0}\right)=0 $$ since \(y=0\) is a solution of this initial value problem, the uniqueness part of Theorem 4. 1. I yields a contradiction. Thus \(W\) is never zero.

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ \begin{array}{l}{2 y^{\mathrm{iv}}-y^{\prime \prime \prime}+4 y^{\prime}+4 y=0 ; \quad y(0)=-2, \quad y^{\prime}(0)=0, \quad y^{\prime \prime}(0)=-2} \\\ {y^{\prime \prime \prime}(0)=0}\end{array} $$

Verify that the differential operator defined by $$ L[y]=y^{(n)}+p_{1}(t) y^{(n-1)}+\cdots+p_{n}(t) y $$ is a linear differential operator. That is, show that $$ L\left[c_{1} y_{1}+c_{2} y_{2}\right]=c_{1} L\left[y_{1}\right]+c_{2} L\left[y_{2}\right] $$ where \(y_{1}\) and \(y_{2}\) are \(n\) times differentiable functions and \(c_{1}\) and \(c_{2}\) are arbitrary constants. Hence, show that if \(y_{1}, y_{2}, \ldots, y_{n}\) are solutions of \(L[y]=0,\) then the linear combination \(c_{1} y_{1}+\cdots+c_{n} y_{n}\) is also a solution of \(L[y]=0 .\)

Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ \sqrt{3}-i $$

determine intervals in which solutions are sure to exist. $$ \left(x^{2}-4\right) y^{\mathrm{vi}}+x^{2} y^{\prime \prime \prime}+9 y=0 $$

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