/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 From \(m^{2}+4=0\) we find \(m_{... [FREE SOLUTION] | 91Ó°ÊÓ

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From \(m^{2}+4=0\) we find \(m_{1}=2 i\) and \(m_{2}=-2 i .\) Then \(y_{c}=c_{1} \cos 2 x+c_{2} \sin 2 x\) and we assume \(y_{p}=A x \cos 2 x+B x \sin 2 x .\) Substituting into the differential equation we obtain \(4 B=0\) and \(-4 A=3\) Then \(A=-\frac{3}{4}, B=0, y_{p}=-\frac{3}{4} x \cos 2 x,\) and \(y=c_{1} \cos 2 x+c_{2} \sin 2 x-\frac{3}{4} x \cos 2 x\).

Short Answer

Expert verified
The full solution is \(y = c_1 \cos(2x) + c_2 \sin(2x) - \frac{3}{4}x \cos(2x)\).

Step by step solution

01

Identify Roots and General Solution

We find the roots from the equation \(m^2 + 4 = 0\), which are \(m_1 = 2i\) and \(m_2 = -2i\). This indicates a complex conjugate pair, suggesting a sinusoidal phenomenon. Therefore, the complementary solution (general solution for the homogeneous equation) is\[ y_c = c_1 \cos(2x) + c_2 \sin(2x) \]where \(c_1\) and \(c_2\) are constants of integration.
02

Assume Specific Solution Form

The form for the particular solution \(y_p\) is assumed based on the need for a non-homogeneous term:\[ y_p = Ax \cos(2x) + Bx \sin(2x) \]This form is chosen because the particular integral follows the same frequency as the complementary solution, but is multiplied by \(x\) to avoid duplication of terms.
03

Substitute and Match Coefficients

We substitute \(y_p = Ax \cos(2x) + Bx \sin(2x)\) into the differential equation. By matching the coefficients, we get two separate equations: \[ 4B = 0 \]and\[ -4A = 3 \]Solving these equations gives us the values for \(A\) and \(B\).
04

Solve for A and B

From \(4B = 0\), we find \(B = 0\). From \(-4A = 3\), we find \(A = -\frac{3}{4}\). Therefore, the particular solution is:\[ y_p = -\frac{3}{4}x \cos(2x) \]
05

Combine General and Particular Solutions

The complete solution to the differential equation is the sum of the complementary solution and particular solution:\[ y = y_c + y_p = c_1 \cos(2x) + c_2 \sin(2x) - \frac{3}{4}x \cos(2x) \]

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Key Concepts

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

Complex Roots
In the world of differential equations, encountering complex roots is not uncommon, especially in second-order linear equations. When we solve the characteristic equation like \( m^2 + 4 = 0 \), the result is a pair of complex conjugate roots, \( m_1 = 2i \) and \( m_2 = -2i \). Complex roots occur when the discriminant (the part under the square root in the quadratic formula) is negative, indicating that no real roots exist.

Complex roots are important because they indicate oscillatory solutions, which are common in systems modeling harmonic oscillations, such as in electrical circuits or mechanical vibrations. Instead of exponential growth or decay seen with real roots, the solution will involve sinusoidal functions like sine and cosine. This connection between complex roots and sinusoidal functions is foundational in understanding oscillating systems.
Complementary Solution
The complementary, or homogeneous, solution of a differential equation involves solving the equation assuming no external force or input. In our case, with complex roots obtained earlier, the complementary solution takes the form of sine and cosine functions. Specifically, the complementary solution derived from the roots \( m_1 = 2i \) and \( m_2 = -2i \) is:
  • \( y_c = c_1 \cos(2x) + c_2 \sin(2x) \)
Here, \( c_1 \) and \( c_2 \) are constants determined by initial conditions. These constants allow the solution to be adjusted to fit the particular conditions of a problem, such as initial velocity or position in a mechanical system.

This solution represents the natural behavior of the system without any external influences, showcasing how the system inherently oscillates due to its properties.
Particular Solution
To solve a non-homogeneous differential equation, one must find a particular solution that accounts for external forces or inputs. The presence of terms similar to those in the complementary solution often leads to the particular solution being multiplied by \( x \) to distinguish it and avoid redundancy.

In this instance, we preassumed a particular solution with the same frequency components:
  • \( y_p = Ax\cos(2x) + Bx\sin(2x) \)
By substituting back into the differential equation, we were able to determine the constants \( A \) and \( B \) through a process called "matching coefficients," leading to:
  • \( A = -\frac{3}{4} \)
  • \( B = 0 \)
Ultimately, the specific external force described by the differential equation shaped this particular solution:
  • \( y_p = -\frac{3}{4}x \cos(2x) \)
Sinusoidal Solutions
Sinusoidal solutions arise when differential equations feature complex roots, as they directly map to the oscillatory nature of sine and cosine waves. These solutions are characterized by frequencies and amplitudes that can describe everything from sound waves to electrical signals.
  • The sinusoidal complementary solution is:\( y_c = c_1 \cos(2x) + c_2 \sin(2x) \)
  • The sinusoidal particular solution is:\( y_p = -\frac{3}{4}x \cos(2x) \)
When these components are combined, they form a complete solution that accounts for both the system's inherent oscillation and any external forcing terms:
  • \( y = y_c + y_p = c_1 \cos(2x) + c_2 \sin(2x) - \frac{3}{4}x \cos(2x) \)
This blending of sinusoidal functions is crucial in understanding and predicting the behavior of systems under various influences, showcasing the elegant harmony between natural oscillations and driven responses.

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Most popular questions from this chapter

Using a CAS to solve the auxiliary equation \(3.15 m^{4}-5.34 m^{2}+6.33 m-2.03=0\) we find \(m_{1}=-1.74806\) \(m_{2}=0.501219, m_{3}=0.62342+0.588965 i,\) and \(m_{4}=0.62342-0.588965 i .\) The general solution is $$y=c_{1} e^{-1.74806 x}+c_{2} e^{0.501219 x}+e^{0.62342 x}\left(c_{3} \cos 0.588965 x+c_{4} \sin 0.588965 x\right)$$

The auxiliary equation is \(m^{3}+m=m\left(m^{2}+1\right)=0,\) so \(y_{c}=c_{1}+c_{2} \cos x+c_{3} \sin x\) and $$W=\left|\begin{array}{rrr}1 & \cos x & \sin x \\\0 & -\sin x & \cos x \\ 0 & -\cos x & -\sin x\end{array}\right|=1$$ Identifying \(f(x)=\tan x\) we obtain $$u_{1}^{\prime}=W_{1}=\left|\begin{array}{ccr}0 & \cos x & \sin x \\\0 & -\sin x & \cos x \\ \tan x & -\cos x & -\sin x\end{array}\right|=\tan x$$ $$u_{2}^{\prime}=W_{2}=\left|\begin{array}{rrr}1 & 0 & \sin x \\\0 & 0 & \cos x \\\0 & \tan x & -\sin x\end{array}\right|=-\sin x$$ $$u_{3}^{\prime}=W_{3}=\left|\begin{array}{ccc}1 & \cos x & 0 \\\0 & -\sin x & 0 \\\0 & -\cos x & \tan x\end{array}\right|=-\sin x \tan x=\frac{\cos ^{2} x-1}{\cos x}=\cos x-\sec x$$ Then $$\begin{array}{l}u_{1}=-\ln |\cos x| \\\u_{2}=\cos x \\\u_{3}=\sin x-\ln |\sec x+\tan x| \end{array}$$ and $$y=c_{1}+c_{2} \cos x+c_{3} \sin x-\ln |\cos x|+\cos ^{2} x$$ $$+\sin ^{2} x-\sin x \ln |\sec x+\tan x|$$ $$=c_{4}+c_{2} \cos x+c_{3} \sin x-\ln |\cos x|-\sin x \ln |\sec x+\tan x|$$ for \(-\pi / 2

From \(k_{1}=40\) and \(k_{2}=120\) we compute the effective spring constant \(k=4(40)(120) / 160=120 .\) Now, \(m=20 / 32\) so \(k / m=120(32) / 20=192\) and \(x^{\prime \prime}+192 x=0 . \quad\) Using \(x(0)=0\) and \(x^{\prime}(0)=2\) we obtain \(x(t)=\frac{\sqrt{3}}{12} \sin 8 \sqrt{3} t\).

We see from the graph that \(\tan x=-x\) has infinitely many roots. since \(\lambda_{n}=\alpha_{n}^{2},\) there are no new eigenvalues when \(\alpha_{n}<0 .\) For \(\lambda=0,\) the differential equation \(y^{\prime \prime}=0\) has general solution \(y=c_{1} x+c_{2}\). The boundary conditions imply \(c_{1}=c_{2}=0,\) so \(y=0\).

(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.

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