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Chapter 5: Problem 9

In Exercises \(1-12\) find the general solution. $$ y^{\prime \prime}-2 y^{\prime}+3 y=0 $$

Short Answer

Expert verified
Answer: The general solution is \(y(x) = e^x(C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x))\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step by step solution

01

Find the characteristic equation

To find the characteristic equation, replace \(y^{\prime \prime}\) with \(m^2\), \(y^{\prime}\) with \(m\), and \(y\) with \(1\). This gives us the characteristic equation: $$ m^2 - 2m + 3 = 0 $$
02

Solve the characteristic equation for the roots

Now, solve the characteristic equation for the roots: $$ m^2 - 2m + 3 = 0 $$ As this is a quadratic equation, we can use the quadratic formula to find the roots: $$ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ Here, \(a = 1\), \(b = -2\), \(c = 3\). Plug these values into the formula to find the roots of the characteristic equation: $$ m = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(3)}}{2(1)} $$ Upon calculating, we get: $$ m = 1 \pm i\sqrt{2} $$ So, the roots are complex conjugate, which are \(m_1 = 1 + i\sqrt{2}\) and \(m_2 = 1 - i\sqrt{2}\).
03

Construct the general solution using the roots

Since the roots are complex conjugate, the general solution for the differential equation will be in the following form: $$ y(x) = e^{mx}(C_1 \cos(\omega x) + C_2 \sin(\omega x)) $$ Where \(m\) is the real part of the root, and \(\omega\) is the imaginary part of the root. Using the roots obtained in step 2, we have \(m = 1\) and \(\omega = \sqrt{2}\). Finally, plug these values into the general solution formula: $$ y(x) = e^x(C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x)) $$ Thus, the general solution for the given differential equation is: $$ y(x) = e^x(C_1 \cos(\sqrt{2}x) + C_2 \sin(\sqrt{2}x)) $$

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Key Concepts

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

Characteristic Equation
In the world of differential equations, the characteristic equation plays a crucial role, especially when dealing with linear differential equations with constant coefficients. To find it, we use a simple yet effective method known as the auxiliary equation approach. We begin by replacing all derivatives in the given differential equation with powers of a variable, typically denoted as \( m \).
For instance, in the differential equation \( y'' - 2y' + 3y = 0 \), replace \( y'' \) with \( m^2 \), \( y' \) with \( m \), and \( y \) with \( 1 \).
This yields the characteristic equation: \( m^2 - 2m + 3 = 0 \). By equating this to zero, we transform the differential equation into an algebraic problem, allowing us to solve for \( m \), which will then inform the structure of the general solution.
  • The power of this method lies in its effectiveness in reducing complex differential equations to simpler algebraic forms.
  • Once we find the roots of this characteristic equation, we can proceed to construct the general solution.
Complex Conjugate Roots
When dealing with second-order differential equations, the nature of the roots of the characteristic equation greatly influences the form of the general solution. In the case where the roots are complex conjugates, such as in our equation \( m^2 - 2m + 3 = 0 \), we get roots like \( 1 + i\sqrt{2} \) and \( 1 - i\sqrt{2} \).
These are called complex conjugate roots because they take the form \( a \pm bi \). Here \( a = 1 \) is the real part and \( b = \sqrt{2} \) is the imaginary part.
  • Complex conjugate roots imply oscillatory solutions in the context of differential equations.
  • The presence of an imaginary part in the roots indicates that the solution will involve sinusoidal functions: sine and cosine.
Understanding these roots is vital as they determine whether the solution involves exponential growth, decay, or oscillations.
General Solution
The general solution of a differential equation encompasses all possible solutions that satisfy the given differential equation. For an equation with complex conjugate roots, the general solution takes on a specific form.
For the roots \( 1 + i\sqrt{2} \) and \( 1 - i\sqrt{2} \), the solution is represented as: \[ y(x) = e^{ax}(C_1 \cos(bx) + C_2 \sin(bx)) \]Here, \( a = 1 \) (the real part of the roots) and \( b = \sqrt{2} \) (the magnitude of the imaginary component).
  • The term \( e^{ax} \) denotes an exponential function which, due to the real part \( a \), indicates exponential growth or decay.
  • Combined with the oscillatory \( \cos(bx) \) and \( \sin(bx) \) components, this form constructs the elegant solution.
  • \( C_1 \) and \( C_2 \) are arbitrary constants, determined by initial conditions or boundary values.
This structure elegantly combines exponential and trigonometric components to capture the full spectrum of behaviors modeled by the differential equation.
Quadratic Formula
The quadratic formula is a powerful mathematical tool used to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). In the context of our problem, it aids in finding the roots of the characteristic equation, \( m^2 - 2m + 3 = 0 \), which is a classic quadratic equation.
The formula is expressed as: \[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]By plugging in our coefficients \( a = 1 \), \( b = -2 \), and \( c = 3 \) into the formula, we proceed as follows:
  • Calculate the discriminant, \( b^2 - 4ac \). For our equation, it is \((-2)^2 - 4(1)(3) = 4 - 12 = -8 \).
  • The negative discriminant indicates the presence of complex roots.
  • Solving further, we obtain \( m = 1 \pm i\sqrt{2} \), aligning with the roots derived from the characteristic equation.
The quadratic formula not only reveals the nature of the roots (real or complex) through the discriminant but also precisely computes those roots, paving the way to forming the general solution.

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Most popular questions from this chapter

Use the principle of superposition to find a particular solution. Where indicated, solve the initial value problem. $$ y^{\prime \prime}-4 y^{\prime}+4 y=6 e^{2 x}+25 \sin x, \quad y(0)=5, y^{\prime}(0)=3 $$

Use variation of parameters to find a particular solution, given the solutions \(y_{1}, y_{2}\) of the complementary equation. $$ 4 x^{2} y^{\prime \prime}-4 x y^{\prime}+\left(3-16 x^{2}\right) y=8 x^{5 / 2} ; \quad y_{1}=\sqrt{x} e^{2 x}, y_{2}=\sqrt{x} e^{-2 x} $$

In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented by the infinite series $$ \begin{array}{c} e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2 !}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\ \cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2 !}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots, \end{array} $$ and $$ \sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}+\cdots $$ for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real, to obtain $$ e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} $$ Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\). (a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence \(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive: $$ 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots $$ Use this to group the terms in (D) as $$ \begin{aligned} e^{i \theta} &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$ In calculus you learned that \(e^{u}, \cos u,\) and \(\sin u\) can be represented by the infinite series $$ \begin{array}{c} e^{u}=\sum_{n=0}^{\infty} \frac{u^{n}}{n !}=1+\frac{u}{1 !}+\frac{u^{2}}{2 !}+\frac{u^{3}}{3 !}+\cdots+\frac{u^{n}}{n !}+\cdots \\ \cos u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n}}{(2 n) !}=1-\frac{u^{2}}{2 !}+\frac{u^{4}}{4 !}+\cdots+(-1)^{n} \frac{u^{2 n}}{(2 n) !}+\cdots, \end{array} $$ and $$ \sin u=\sum_{n=0}^{\infty}(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}=u-\frac{u^{3}}{3 !}+\frac{u^{5}}{5 !}+\cdots+(-1)^{n} \frac{u^{2 n+1}}{(2 n+1) !}+\cdots $$ for all real values of \(u\). Even though you have previously considered (A) only for real values of \(u\), we can set \(u=i \theta,\) where \(\theta\) is real, to obtain $$ e^{i \theta}=\sum_{n=0}^{\infty} \frac{(i \theta)^{n}}{n !} $$ Given the proper background in the theory of infinite series with complex terms, it can be shown that the series in (D) converges for all real \(\theta\). (a) Recalling that \(i^{2}=-1,\) write enough terms of the sequence \(\left\\{i^{n}\right\\}\) to convince yourself that the sequence is repetitive: $$ 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, 1, i,-1,-i, \cdots $$ Use this to group the terms in (D) as $$ \begin{aligned} e^{i \theta} &=\left(1-\frac{\theta^{2}}{2}+\frac{\theta^{4}}{4}+\cdots\right)+i\left(\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}+\cdots\right) \\ &=\sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n}}{(2 n) !}+i \sum_{n=0}^{\infty}(-1)^{n} \frac{\theta^{2 n+1}}{(2 n+1) !} \end{aligned} $$ (c) If \(\alpha\) and \(\beta\) are real numbers, define $$ e^{\alpha+i \beta}=e^{\alpha} e^{i \beta}=e^{\alpha}(\cos \beta+i \sin \beta) $$ Show that if \(z_{1}=\alpha_{1}+i \beta_{1}\) and \(z_{2}=\alpha_{2}+i \beta_{2}\) then $$ e^{z_{1}+z_{2}}=e^{z_{1}} e^{z_{2}} $$ (d) Let \(a, b,\) and \(c\) be real numbers, with \(a \neq 0 .\) Let \(z=u+i v\) where \(u\) and \(v\) are real-valued functions of \(x .\) Then we say that \(z\) is a solution of $$ a y^{\prime \prime}+b y^{\prime}+c y=0 $$ if \(u\) and \(v\) are both solutions of (G). Use Theorem \(5.2 .1(\mathbf{c})\) to verify that if the characteristic equation of (G) has complex conjugate roots \(\lambda \pm i \omega\) then \(z_{1}=e^{(\lambda+i \omega) x}\) and \(z_{2}=e^{(\lambda-i \omega) x}\) are both solutions of (G).

The nonlinear first order equation $$y^{\prime}+y^{2}+p(x) y+q(x)=0$$ is a Riccati equation. (See Exercise 2.4.55.) Assume that \(p\) and \(q\) are continuous. (a) Show that \(y\) is a solution of \((\mathrm{A})\) if and only if \(y=z^{\prime} / z,\) where $$z^{\prime \prime}+p(x) z^{\prime}+q(x) z=0$$ (b) Show that the general solution of \((\mathrm{A})\) is $$y=\frac{c_{1} z_{1}^{\prime}+c_{2} z_{2}^{\prime}}{c_{1} z_{1}+c_{2} z_{2}}$$ where \(\left\\{z_{1}, z_{2}\right\\}\) is a fundamental set of solutions of (B) and \(c_{1}\) and \(c_{2}\) are arbitrary constants. (c) Does the formula (C) imply that the first order equation (A) has a two- parameter family of solutions? Explain your answer.

In Exercises \(1-6\) use variation of parameters to find a particular solution. $$ y^{\prime \prime}+9 y=\tan 3 x $$
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