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Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}-4 y^{\prime \prime \prime}+4 y^{\prime \prime}=0 $$

Short Answer

Expert verified
Answer: The general solution of the given differential equation is \(y(x) = C_1 + C_2x + C_3e^{2x} + C_4xe^{2x}\), where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are arbitrary constants.

Step by step solution

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01

Find the characteristic equation

Replace the given derivatives with the power of variable \(r\). The characteristic equation is: $$ r^4 - 4r^3 + 4r^2 = 0 $$
02

Solve the characteristic equation

Factor the characteristic equation to find the values of \(r\): $$ r^2 (r^2 - 4r + 4) = 0 $$ We can further simplify it: $$ r^2 (r - 2)^2 = 0 $$ Now, we have found the root of the characteristic equation: \(r_1 = 0\), \(r_2 = 0\), \(r_3 = 2\), \(r_4 = 2\).
03

Write the general solution

Using the values of \(r\), we can write the general solution for the given differential equation in the form of: $$ y(x) = C_1e^{0x} + C_2xe^{0x} + C_3e^{2x} + C_4xe^{2x} $$ Simplifying it further, we get the general solution: $$ y(x) = C_1 + C_2x + C_3e^{2x} + C_4xe^{2x} $$ Where \(C_1\), \(C_2\), \(C_3\), and \(C_4\) are arbitrary constants.

Key Concepts

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

Characteristic Equation
Understanding the characteristic equation is fundamental when solving linear homogeneous differential equations with constant coefficients. This mathematical tool translates the problem of finding solutions to the differential equation into finding the roots of a polynomial, which is, in most cases, a simpler problem to solve.

The process begins by assuming a solution for the differential equation in the form of an exponential function, typically denoted by \( e^{rx} \), where \( r \) is a constant. Corresponding derivatives of this assumed solution are then substituted back into the original differential equation, resulting in a polynomial equation in terms of \( r \), known as the characteristic equation.

For example, in the exercise given, replacing derivatives with powers of \( r \) in the equation \( y^{\text{iv}} - 4y''' + 4y'' = 0 \) leads to the polynomial \( r^4 - 4r^3 + 4r^2 = 0 \). Solving this equation reveals the roots that correspond to the constants in the exponent of the exponential solutions to the differential equation. It acts as a bridge between the differential equation and its solutions.
Homogeneous Differential Equation
A homogeneous differential equation is a specific type of differential equation that equates to zero and contains derivatives of a function, but not the function itself. This property ensures that the sum of any two solutions is also a solution - a principle known as the superposition principle.

In the context of the provided problem, the differential equation \( y^{\text{iv}} - 4y''' + 4y'' = 0 \) is homogeneous because it does not have any terms that are not derivatives of \( y \). The superposition principle is crucial as it implies that the general solution can be expressed as a linear combination of solutions corresponding to each root of the characteristic equation. This brings us to the beautifully simple yet powerful exponential form of the solution that incorporates arbitrary constants, reflecting the infinite number of solutions typical for homogeneous equations.

The general form of a solution to a homogeneous differential equation with constant coefficients, as visible in the exercise, includes exponential functions where the exponents contain the roots of the characteristic equation. The coefficients of these exponential terms are the arbitrary constants we will discuss next.
Arbitrary Constants
Arbitrary constants in the solution of a differential equation represent the general nature of the solution, encompassing a whole family of specific solutions. Each specific solution within this family is determined by initial conditions or boundary values that are often dictated by the context of a problem.

In the provided exercise, the general solution \( y(x) = C_1 + C_2x + C_3e^{2x} + C_4xe^{2x} \) includes four arbitrary constants: \( C_1, C_2, C_3, \) and \( C_4 \). These constants take on values that fulfill the particular requirements of a real-world scenario, experimental data, or initial/boundary conditions provided in the problem.

For instance, if initial values of the function and several derivatives at a point are known, these constants can be uniquely determined, and the previously 'general' solution becomes a 'particular' solution. Without specific conditions, these constants remain arbitrary, highlighting another aspect of the beauty of mathematics - its ability to abstract and generalize across infinite possibilities.

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Most popular questions from this chapter

Find the general solution of the given differential equation. $$ y^{\mathrm{iv}}+2 y^{\prime \prime}+y=0 $$

verify that the given functions are solutions of the differential equation, and determine their Wronskian. $$ y^{w}+2 y^{\prime \prime \prime}+y^{\prime \prime}=0 ; \quad 1, \quad t ; \quad e^{-t}, \quad t e^{-t} $$

Consider the nonhomogeneous \(n\) th order linear differential equation $$ a_{0} y^{(n)}+a_{1} y^{(n-1)}+\cdots+a_{n} y=g(t) $$ where \(a_{0}, \ldots, a_{n}\) are constants. Verify that if \(g(t)\) is of the form $$ e^{\alpha t}\left(b_{0} t^{m}+\cdots+b_{m}\right) $$ then the substitution \(y=e^{\alpha t} u(t)\) reduces the preceding equation to the form $$ k_{0} u^{(n)}+k_{1} u^{(n-1)}+\cdots+k_{n} u=b_{0} t^{m}+\cdots+b_{m} $$ where \(k_{0}, \ldots, k_{n}\) are constants. Determine \(k_{0}\) and \(k_{n}\) in terms of the \(a^{\prime}\) 's and \(\alpha .\) Thus the problem of determining a particular solution of the original equation is reduced to the simpler problem of determining a particular solution of an equation with constant coefficients and a polynomial for the nonhomogeneous term.

Find the solution of the given initial value problem and plot its graph. How does the solution behave as \(t \rightarrow \infty ?\) $$ y^{\mathrm{iv}}-4 y^{\prime \prime \prime}+4 y^{\prime \prime}=0 ; \quad y(1)=-1, \quad y^{\prime}(1)=2, \quad y^{\prime \prime}(1)=0, \quad y^{\prime \prime \prime}(1)=0 $$

Follow the procedure illustrated in Example 4 to determine the indicated roots of the given complex number. $$ [2(\cos \pi / 3+i \sin \pi / 3)]^{1 / 2} $$

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