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Chapter 8: Problem 9

Solve the following differential equations. \(\left(D^{2}-4 D+13\right) y=0\)

Short Answer

Expert verified
The solution is \[y = e^{2x} \left(C_1 \cos(3x) + C_2 \sin(3x) \right)\].

Step by step solution

01

Identify the Differential Operator

Recognize that the given differential equation \(\left(D^{2}-4 D+13\right) y=0\)can be approached by analyzing the characteristic equation. The differential operator here is \(D^2 - 4D + 13\).
02

Form the Characteristic Equation

Form the characteristic equation from the differential operator: \(r^2 - 4r + 13 = 0\). This is a quadratic equation.
03

Solve the Characteristic Equation

Solve the quadratic equation: \(r^2 - 4r + 13 = 0\). We use the quadratic formula \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\], where \(a = 1\), \(b = -4\), and \(c = 13\).
04

Calculate the Roots

Substitute the values into the quadratic formula: \[r = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i.\]The roots are \(2 + 3i\) and \(2 - 3i\).
05

Form the General Solution

Since the roots are complex, the general solution for the differential equation is \[y = e^{\alpha x} \left(C_1 \cos(\beta x) + C_2 \sin(\beta x) \right)\], where \(\alpha = 2\) and \(\beta = 3\).Therefore, the general solution is \[y = e^{2x} \left(C_1 \cos(3x) + C_2 \sin(3x) \right)\].

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Key Concepts

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

Characteristic Equation
A key step in solving second order differential equations is to form the characteristic equation. When presented with a differential equation like \((D^{2}-4 D+13)y=0\), recognize that 'D' represents the differential operator. The characteristic equation is formed directly from the given differential operator by replacing D with 'r'. This results in the characteristic equation: \[r^2 - 4r + 13 = 0.\] This quadratic equation is crucial for determining the nature of the roots, which in turn informs the form of the general solution.
Complex Roots
Once the characteristic equation is formed, solving it using the quadratic formula gives us the roots. If the discriminant (\[b^2 - 4ac\]) is negative, the roots are complex. In our example, the characteristic equation is \[r^2 - 4r + 13 = 0,\] with \[a = 1,\ b = -4,\] and \[c = 13.\] Calculating the discriminant, \[16 - 52 = -36,\] we find it to be negative, indicating complex roots. These roots are computed as \[r = \frac{4 \pm 6i}{2} = 2 \pm 3i.\] Hence, the roots are \[2 + 3i\] and \[2 - 3i.\]
General Solution
For second order differential equations with complex roots, the general solution has a specific form. Given roots \[2 + 3i\] and \[2 - 3i\], we use the formula for the general solution: \[y = e^{\alpha x} \left(C_1 \cos(\beta x) + C_2 \sin(\beta x) \right),\] where \[\alpha\] is the real part of the root, and \[\beta\] is the imaginary part. In this case, \[\alpha = 2\] and \[\beta = 3.\] This gives the general solution: \[y = e^{2x} \left(C_1 \cos(3x) + C_2 \sin(3x) \right).\] The constants \[C_1\] and \[C_2\] are determined from initial conditions if provided.
Quadratic Formula
The quadratic formula is essential for finding the roots of the characteristic equation. It is given by: \[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\] For our equation, \[r^2 - 4r + 13 = 0,\] with \[a = 1,\ b = -4,\ c = 13,\] substituting these values we get: \[r = \frac{4 \pm \sqrt{-36}}{2} = \frac{4 \pm 6i}{2} = 2 \pm 3i.\] This formula allows us to handle both real and complex cases effectively, providing a complete solution method for the characteristic equations.
Exponential and Trigonometric Functions
In the context of solving differential equations with complex roots, the general solution often involves exponential and trigonometric functions. This stems from Euler's formula: \[e^{i\beta x} = \cos(\beta x) + i\sin(\beta x).\] Here, the exponential part \[e^{\alpha x}\] comes from the real part of the root \[\alpha,\] while the trigonometric parts \[\cos(\beta x)\] and \[\sin(\beta x)\] stem from the imaginary part \[\beta.\] These functions combine to form a solution that describes oscillatory behavior modulated by an exponential growth or decay, reflecting the complex root nature in simple harmonic motion or wave-like patterns with exponential modifiers.

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Most popular questions from this chapter

Solve the following differential equations. \(y^{\prime \prime}+y^{\prime}-2 y=0\)

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

The momentum \(p\) of an electron at velocity \(v\) near the velocity \(c\) of light increases according to the formula $$ p=\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}} $$ where \(m_{0}\) is a constant (the rest mass). If an electron is subject to a constant force \(F\), Newton's second law describing its motion is $$ \frac{d p}{d t}=\frac{d}{d t}\left(\frac{m_{0} v}{\sqrt{1-v^{2} / c^{2}}}\right)=F $$ Find the velocity as a function of time and show that the limiting velocity as \(t\) tends to infinity is \(c\). Find the distance traveled by the electron in time \(t\) if it starts from rest.

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(y^{\prime \prime}+4 y^{\prime}+5 y=2 e^{-2 x} \cos x\)

By separation of variables, solve the differential equation \(d y / d x=\sqrt{1-y^{2}}\) to obtain solution containing one arbitrary constant. Although this solution may be referred to as the " general solution," show that \(y=1\) is a solution of the differential equation not obtainablc from the "general solution" by any choice of the arbitrary constant. The solution \(y=1\) is called a singular selution; \(y=-1\) is another singular solution. Sketch a number of graphs of the "general solution" for different values of the arbitrary constant and observe that \(y=1\) is tangent to all of them. This is characteristic of a singular solution -its graph is tangent at each point to one of the graphs of the "general solution." Note that the given differential equation is not linear; for linear equations, all solutions are contained in the general solution, but nonlinear equations may have singular solutions which cannot be obtained from the "general solution" by specializing the arbitrary constant (or constants). Thus a nonlinear first-order equation in \(x\) and \(y\) may have two (or more) solutions passing through a given point in the \((x, y)\) plane, whereas a linear first-order equation always has just one such solution. Show that any continuous curve made up of pieces of \(y=1, y=-1\), and the sinc curves of the "general solution," gives a solution of the above differential equation. Sketch such a solution curve on your graphs.
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