Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \((D-2)^{2}\left(D^{2}+9\right) y=0\)

Short answer

Linear fourth-order solution: \( y = \left(C1 + C2x\right)e^{2x} + C3 \cos(3x) + C4 \sin(3x) \).

Step-by-step solution

- Identify the Type of Differential Equation

First, recognize that \( (D-2)^{2}\big(D^{2}+9\big) y=0 \) is a linear homogeneous differential equation with constant coefficients. Here, \( D \) represents the differentiation operator \( \frac{d}{dx} \).

- Find the Characteristic Equation

Form the characteristic equation by replacing \( D \) with \( r \) in the given differential operator. The characteristic equation is: \i.e., \( (r-2)^{2}(r^{2}+9) = 0 \).

- Solve the Characteristic Equation

Solve \( (r-2)^{2}(r^{2}+9) = 0 \). This gives the repeated root \( r=2 \) with multiplicity 2, and the complex roots \( r= \pm 3i \).

- Write the General Solution

Use the roots to write the general solution of the differential equation. Since \( r=2 \) is a repeated root, and \( r= \pm 3i \) are complex roots, the general solution is: \y = \( (C1 + C2x)e^{2x} + C3 \cos(3x) + C4 \sin(3x) \) where \( C1, C2, C3, \, \, \, and \, C4 \) are arbitrary constants.

Key concepts

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In the exercise, we encountered a **linear homogeneous differential equation**. This is a differential equation in which every term depends on the unknown function and its derivatives, and importantly, the equation equals zero. Such equations are crucial in understanding many physical systems where the principle of superposition applies. Another key feature is that they have constant coefficients, meaning coefficients are numbers rather than functions of the independent variable. This simplifies solving them, as the differential operator can be treated algebraically using characteristic equations.

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A related concept is the **characteristic equation**. To form this equation, you simply replace the differentiation operator \(\frac{d}{dx}\) with an algebraic variable, usually represented by \(r\). For our problem, the differential equation \((D-2)^{2}\big(D^{2}+9\big) y=0\) becomes the characteristic equation \((r-2)^{2}( r^2+9)=0\). This equation helps us find the roots, which are essential in identifying the general solution of the differential equation.

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Next, we explored **roots and the general solution**. Solving the characteristic equation \((r-2)^{2}(r^2+9)=0\) gives us the roots \r=2\ (a repeated root), and \(\pm 3i\) (complex roots). These roots are crucial in constructing the general solution. For real repeated roots like \r=2\, we use forms such as \(C_1 e^{2x}+C_2xe^{2x}\). For complex roots like \pm 3i\, we use combinations of sine and cosine: \(C_3 \cos(3x)+ C_4 \sin(3x)\). The general solution combines these forms, resulting in \(y = (C1 + C2x)e^{2x} + C3 \cos(3x) + C4 \sin( 3x)\).

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Finally, **constant coefficients** play a significant role. These coefficients in the differential equation do not vary with the independent variable, making the characteristic equation straightforward to handle. Mathematically, they ensure that the solution techniques are algebraic and facilitate the use of exponentials and trigonometric functions for solutions. Constant coefficients are foundational to these types of problems because they simplify finding and interpreting the solution.