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Chapter 3: Problem 46

The auxiliary equation is \(m(m-1)(m-2)(m-3)-6 m(m-1)(m-2)+33 m(m-1)-105 m+169=0,\) so that \(m_{1}=m_{2}=3+2 i\) and \(m_{3}=m_{4}=3-2 i .\) The general solution of the differential equation is $$y=x^{3}\left[c_{1} \cos (2 \ln x)+c_{2} \sin (2 \ln x)\right]+x^{3} \ln x\left[c_{3} \cos (2 \ln x)+c_{4} \sin (2 \ln x)\right].$$

Short Answer

Expert verified
The general solution is given correctly in the problem statement.

Step by step solution

01

Identify the Roots

The auxiliary equation is given as \( m(m-1)(m-2)(m-3)-6 m(m-1)(m-2)+33 m(m-1)-105 m+169=0 \). You are told that \( m_{1}=m_{2}=3+2i \) and \( m_{3}=m_{4}=3-2i \). These are conjugate pairs with double roots.
02

Write the General Solution Structure

For each pair of complex roots \( m = a + bi \) and \( m = a - bi \) with multiplicity, the corresponding part of the solution is given by \( x^a [c_1 \cos(b \ln x) + c_2 \sin(b \ln x)] + x^a \ln x [c_3 \cos(b \ln x) + c_4 \sin(b \ln x)]. \)
03

Apply the Roots to the Equation

With \( m_{1}=m_{2}=3+2i \) and \( m_{3}=m_{4}=3-2i \), our general solution will take the form: \[ y = x^3 [c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] + x^3 \ln x [c_3 \cos(2 \ln x) + c_4 \sin(2 \ln x)]. \]
04

Compile the General Solution

Putting it all together, the general solution to the differential equation is: \[ y = x^3 \left[c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)\right] + x^3 \ln x \left[c_3 \cos(2 \ln x) + c_4 \sin(2 \ln x)\right]. \] This accounts for both conjugate pairs and their multiplicities.

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Key Concepts

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

Complex Roots
When solving differential equations, especially linear ones with constant coefficients, complex roots can arise. These roots are part of the characteristic, or auxiliary, equation, which is derived from the differential equation. In our context, the equation we have is actually an auxiliary polynomial that has complex roots.

**What Are Complex Roots?**
  • In general, a complex number has the form \( a + bi \), where \( a \) and \( b \) are real numbers, and \( i \) is the imaginary unit satisfying \( i^2 = -1 \).
  • A complex root implies there exists a solution to the differential equation that will include trigonometric functions like sine and cosine.

**Conjugate Pairs**
  • Complex roots usually appear in conjugate pairs. This means if \( 3 + 2i \) is a root, its conjugate \( 3 - 2i \) will also be a root.
  • This property helps us convert complex solutions into real-valued solutions using Euler's formula.
General Solution
The general solution of a differential equation represents the set of all possible solutions it can have. This solution involves constant coefficients which can be determined by specific initial conditions given in a problem.

**Constructing the General Solution**
  • For complex roots of form \( m = a + bi \), the solution includes terms with logarithmic and trigonometric components: \[y = x^a [c_1 \cos(b \ln x) + c_2 \sin(b \ln x)] + x^a \ln x [c_3 \cos(b \ln x) + c_4 \sin(b \ln x)].\]
  • Here, \( c_1, c_2, c_3, \) and \( c_4 \) are arbitrary constants determined by boundary conditions.

**Application in Our Solution**
  • Given \( m_1 = m_2 = 3 + 2i \) and \( m_3 = m_4 = 3 - 2i \), the general solution becomes \[y = x^3 [c_1 \cos(2 \ln x) + c_2 \sin(2 \ln x)] + x^3 \ln x [c_3 \cos(2 \ln x) + c_4 \sin(2 \ln x)].\]
  • This takes into account the multiplicity of the roots by incorporating \( x^3 \ln x \) terms.
Auxiliary Equation
The auxiliary equation, also known as the characteristic equation, is a polynomial equation derived from a linear differential equation with constant coefficients. It plays a crucial role in finding the roots which correspond to the solutions of the differential equation.

**What is an Auxiliary Equation?**
  • It's a method to transform a differential equation into a solvable algebraic equation.
  • From our differential equation, the auxiliary equation is \[ m(m-1)(m-2)(m-3)-6 m(m-1)(m-2)+33 m(m-1)-105 m+169=0. \]

**Importance of Solving the Auxiliary Equation**
  • The roots (both real and complex) of the auxiliary equation provide the basis for constructing the general solution.
  • Each unique root gives rise to a component of the solution, while repeated roots introduce additional terms such as those involving \( \ln x \) to account for multiplicities.
Solving this equation, you obtain the roots which dictate the form of the solution, completing the connection between the differential equation and its behavior.

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Most popular questions from this chapter

If \(\frac{1}{2} x^{\prime \prime}+\frac{1}{2} x^{\prime}+6 x=10 \cos 3 t, x(0)=-2,\) and \(x^{\prime}(0)=0\) then \\[x_{c}=e^{-t / 2}\left(c_{1} \cos \frac{\sqrt{47}}{2} t+c_{2} \sin \frac{\sqrt{47}}{2} t\right)\\] and \(x_{p}=\frac{10}{3}(\cos 3 t+\sin 3 t)\) so that the equation of motion is \\[x=e^{-t / 2}\left(-\frac{4}{3} \cos \frac{\sqrt{47}}{2} t-\frac{64}{3 \sqrt{47}} \sin \frac{\sqrt{47}}{2} t\right)+\frac{10}{3}(\cos 3 t+\sin 3 t)\\].

Applying integration by parts twice we have $$\begin{aligned} \int e^{a x} f(x) d x &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a} \int e^{a x} f^{\prime}(x) d x \\ &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a}\left[\frac{1}{a} e^{a x} f^{\prime}(x)-\frac{1}{a} \int e^{a x} f^{\prime \prime}(x) d x\right] \\ &=\frac{1}{a} e^{a x} f(x)-\frac{1}{a^{2}} e^{a x} f^{\prime}(x)+\frac{1}{a^{2}} \int e^{a x} f^{\prime \prime}(x) d x \end{aligned}$$ Collecting the integrals we get $$\int e^{a x}\left(f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)\right) d x=\frac{1}{a} e^{a x} f(x)-\frac{1}{a^{2}} e^{a x} f^{\prime}(x)$$ In order for the technique to work we need to have $$\begin{aligned} &\int e^{a x}\left(f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)\right) d x=k \int e^{a x} f(x) d x\\\ &f(x)-\frac{1}{a^{2}} f^{\prime \prime}(x)=k f(x) \end{aligned}$$ where \(k \neq 0 .\) This is the second-order differential equation \(f^{\prime \prime}(x)+a^{2}(k-1) f(x)=0\) If \(k<1, k \neq 0,\) the solution of the differential equation is a pair of exponential functions, in which case the original integrand is an exponential function and does not require integration by parts for its evaluation. Similarly, if \(k=1, f^{\prime \prime}(x)=0\) and \(f(x)\) has the form \(f(x)=a x+b .\) In this case a single application of integration by parts will suffice. Finally, if \(k>1\), the solution of the differential equation is $$f(x)=c_{1} \cos a \sqrt{k-1} x+c_{2} \sin a \sqrt{k-1} x$$ and we see that the technique will work for linear combinations of \(\cos \alpha x\) and \(\sin \alpha x\)

(a) The boundary-value problem is \\[ \frac{d^{4} y}{d x^{4}}+\lambda \frac{d^{2} y}{d x^{2}}=0, \quad y(0)=0, y^{\prime \prime}(0)=0, y(L)=0, y^{\prime}(L)=0 \\] where \(\lambda=\alpha^{2}=P / E I .\) The solution of the differential equation is \(y=c_{1} \cos \alpha x+c_{2} \sin \alpha x+c_{3} x+c_{4}\) and the conditions \(y(0)=0, y^{\prime \prime}(0)=0\) yield \(c_{1}=0\) and \(c_{4}=0 .\) Next, by applying \(y(L)=0, y^{\prime}(L)=0\) to \(y=c_{2} \sin \alpha x+c_{3} x\) we get the system of equations $$\begin{aligned} c_{2} \sin \alpha L+c_{3} L &=0 \\ \alpha c_{2} \cos \alpha L+c_{3} &=0 \end{aligned}$$.To obtain nontrivial solutions \(c_{2}, c_{3},\) we must have the determinant of the coefficients equal to zero: \\[ \left|\begin{array}{rr} \sin \alpha L & L \\ \alpha \cos \alpha L & 1 \end{array}\right|=0 \quad \text { or } \quad \tan \beta=\beta \\] where \(\beta=\alpha L .\) If \(\beta_{n}\) denotes the positive roots of the last equation, then the eigenvalues are found from \(\beta_{n}=\alpha_{n} L=\sqrt{\lambda_{n}} L\) or \(\lambda_{n}=\left(\beta_{n} / L\right)^{2} .\) From \(\lambda=P / E I\) we see that the critical loads are \(P_{n}=\beta_{n}^{2} E I / L^{2}\) With the aid of a CAS we find that the first positive root of \(\tan \beta=\beta\) is (approximately) \(\beta_{1}=4.4934,\) and so the Euler load is (approximately) \(P_{1}=20.1907 E I / L^{2} .\) Finally, if we use \(c_{3}=-c_{2} \alpha \cos \alpha L,\) then the deflection curves are $$y_{n}(x)=c_{2} \sin \alpha_{n} x+c_{3} x=c_{2}\left[\sin \left(\frac{\beta_{n}}{L} x\right)-\left(\frac{\beta_{n}}{L} \cos \beta_{n}\right) x\right]$$ (b) With \(L=1\) and \(c_{2}\) appropriately chosen, the general shape of the first buckling mode, \\[ y_{1}(x)=c_{2}\left[\sin \left(\frac{4.4934}{L} x\right)-\left(\frac{4.4934}{L} \cos (4.4934)\right) x\right] \\] is shown below.

Let \(u=y^{\prime}\) so that \(u^{\prime}=y^{\prime \prime} .\) The equation becomes \(u^{\prime}-(1 / x) u=(1 / x) u^{3},\) which is Bernoulli. Using \(w=u^{-2}\) we obtain \(d w / d x+(2 / x) w=-2 / x .\) An integrating factor is \(x^{2},\) so \\[ \begin{aligned} \frac{d}{d x}\left[x^{2} w\right]=-2 x & \Longrightarrow x^{2} w=-x^{2}+c_{1} \Longrightarrow w=-1+\frac{c_{1}}{x^{2}} \\ & \Longrightarrow u^{-2}=-1+\frac{c_{1}}{x^{2}} \Longrightarrow u=\frac{x}{\sqrt{c_{1}-x^{2}}} \\ & \Longrightarrow \frac{d y}{d x}=\frac{x}{\sqrt{c_{1}-x^{2}}} \Rightarrow y=-\sqrt{c_{1}-x^{2}}+c_{2} \\ & \Longrightarrow c_{1}-x^{2}=\left(c_{2}-y\right)^{2} \Longrightarrow x^{2}+\left(c_{2}-y\right)^{2}=c_{1} \end{aligned} \\]

The thinner curve is obtained using a numerical solver, while the thicker curve is the graph of the Taylor polynomial. We look for a solution of the form \\[ \begin{array}{c} y(x)=y(0)+y^{\prime}(0) x+\frac{1}{2 !} y^{\prime \prime}(0) x^{2}+\frac{1}{3 !} y^{\prime \prime \prime}(0) x^{3}+\frac{1}{4 !} y^{(4)}(0) x^{4} \\ +\frac{1}{5 !} y^{(5)}(0) x^{5}+\frac{1}{6 !} y^{(6)}(0) x^{6} \end{array} \\] From \(y^{\prime \prime}(x)=e^{y}\) we compute \\[ \begin{aligned} y^{\prime \prime \prime}(x) &=e^{y} y^{\prime} \\ y^{(4)}(x) &=e^{y}\left(y^{\prime}\right)^{2}+e^{y} y^{\prime \prime} \\ y^{(5)}(x) &=e^{y}\left(y^{\prime}\right)^{3}+3 e^{y} y^{\prime} y^{\prime \prime}+e^{y} y^{\prime \prime \prime} \\ y^{(6)}(x) &=e^{y}\left(y^{\prime}\right)^{4}+6 e^{y}\left(y^{\prime}\right)^{2} y^{\prime \prime}+3 e^{y}\left(y^{\prime \prime}\right)^{2}+4 e^{y} y^{\prime} y^{\prime \prime \prime}+e^{y} y^{(4)} \end{aligned} \\] Using \(y(0)=0\) and \(y^{\prime}(0)=-1\) we find \\[ y^{\prime \prime}(0)=1, \quad y^{\prime \prime \prime}(0)=-1, \quad y^{(4)}(0)=2, \quad y^{(5)}(0)=-5, \quad y^{(6)}(0)=16 \\] An approximate solution is \\[ y(x)=-x+\frac{1}{2} x^{2}-\frac{1}{6} x^{3}+\frac{1}{12} x^{4}+\frac{1}{24} x^{5}+\frac{1}{45} x^{6} .\\]
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