/*! 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 39 The general solution of the diff... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 39

The general solution of the differential equation \(y^{\prime \prime}+3 y=6 x\) is \(y=c_{1} \cos \sqrt{3} x+c_{2} \sin \sqrt{3} x+2 x .\) The condition \(y(0)=0\) implies \(c_{1}=0\) and so \(y=c_{2} \sin \sqrt{3} x+2 x .\) The condition \(y(1)+y^{\prime}(1)=0\) implies \(c_{2} \sin \sqrt{3}+2+c_{2} \sqrt{3} \cos \sqrt{3}+2=0\) so \(c_{2}=-4 /(\sin \sqrt{3}+\sqrt{3} \cos \sqrt{3}) .\) The solution is $$y=\frac{-4 \sin \sqrt{3} x}{\sin \sqrt{3}+\sqrt{3} \cos \sqrt{3}}+2 x$$

Short Answer

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

Step by step solution

01

Identifying the Homogeneous Solution

The differential equation given is \( y^{\prime \prime} + 3y = 6x \). The corresponding homogeneous equation is \( y^{\prime \prime} + 3y = 0 \). To solve this, we find solutions of the form \( y = e^{rx} \), leading to the characteristic equation \( r^2 + 3 = 0 \). Solving for \( r \) gives \( r = \pm i \sqrt{3} \). Thus, the general solution of the homogeneous equation is \( y_h = c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x) \).
02

Identifying the Particular Solution

To find a particular solution of the non-homogeneous equation, we propose a solution of the form \( y = Ax + B \), where \( A \) and \( B \) are constants to be determined. Substituting this form into the non-homogeneous equation and solving, we find \( A = 2 \), making the particular solution \( y_p = 2x \).
03

Combining Particular and Homogeneous Solutions

The general solution of the differential equation is the sum of the homogeneous and particular solutions: \( y = y_h + y_p = c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x) + 2x \).
04

Applying the Initial Condition \(y(0) = 0\)

Substitute \( x = 0 \) into the general solution: \( y(0) = c_1 \cos(0) + c_2 \sin(0) + 2 \cdot 0 = 0 \), which simplifies to \( c_1 = 0 \). The solution then becomes \( y = c_2 \sin(\sqrt{3}x) + 2x \).
05

Applying the Condition \(y(1) + y^{\prime}(1) = 0\)

First, find the derivative: \( y' = c_2 \sqrt{3} \cos(\sqrt{3}x) + 2 \). At \( x = 1 \), we have \( y(1) = c_2 \sin(\sqrt{3}) + 2 \), and \( y'(1) = c_2 \sqrt{3} \cos(\sqrt{3}) + 2 \). The condition \( y(1) + y'(1) = 0 \) becomes \( c_2 \sin(\sqrt{3}) + 2 + c_2 \sqrt{3} \cos(\sqrt{3}) + 2 = 0 \).
06

Solving for \(c_2\)

The equation \( c_2 [\sin(\sqrt{3}) + \sqrt{3} \cos(\sqrt{3})] + 4 = 0 \) leads to \( c_2 = \frac{-4}{\sin(\sqrt{3}) + \sqrt{3} \cos(\sqrt{3})} \).
07

Writing the Final Solution

Substitute \( c_2 \) back into the expression for \( y \): \( y = \frac{-4 \sin(\sqrt{3}x)}{\sin(\sqrt{3}) + \sqrt{3} \cos(\sqrt{3})} + 2x \). This is the particular solution that satisfies both conditions.

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Key Concepts

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

Homogeneous Solution
The homogeneous solution is an essential part of solving differential equations. It refers to solving the associated homogeneous equation, which is derived by setting the non-homogeneous part (the right side of the equation) to zero. For the given differential equation \(y'' + 3y = 6x\), the homogeneous equation is \(y'' + 3y = 0\). Solving this involves seeking solutions of the form \(y = e^{rx}\), leading to a characteristic equation. The characteristic equation for this problem is: \[r^2 + 3 = 0\] Solving the characteristic equation gives imaginary roots \(r = \pm i \sqrt{3}\). This indicates solutions that are trigonometric functions. Therefore, the homogeneous solution is: \[y_h = c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x)\]Here \(c_1\) and \(c_2\) are arbitrary constants that will be determined later using initial conditions.
Particular Solution
The particular solution addresses the non-homogeneous part of the differential equation. It's the component of the solution that directly solves \(y'' + 3y = 6x\).To find it, assume a simple polynomial form, such as \(y = Ax + B\), where \(A\) and \(B\) are constants to be determined. Substituting \(y = Ax + B\) into the differential equation yields:- The second derivative is zero: \(y'' = 0\)- Substitute into the equation: \(0 + 3(Ax + B) = 6x\) - Simplifying gives: \(3Ax + 3B = 6x\)Comparing coefficients, \(3A = 6\), resulting in \(A = 2\). The constant \(B\) disappears because there's no constant term comparison. Therefore, the particular solution is:\[y_p = 2x\] This particular form accounts for the specific non-homogeneous term in the original equation.
Initial Conditions
To refine the general solution based on specific criteria, we apply initial conditions. These conditions help determine the arbitrary constants in the homogeneous solution.Given the condition \(y(0) = 0\), substitute \(x = 0\) into the general solution \(y = c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x) + 2x\):- \(y(0) = c_1 \cos(0) + c_2 \sin(0) + 2 \cdot 0 = 0\)- Simplifies to \(c_1 = 0\), since \(\cos(0) = 1\) and \(\sin(0) = 0\)This yields a simplified solution:\[y = c_2 \sin(\sqrt{3}x) + 2x\]Another condition \(y(1) + y'(1) = 0\) is used to solve for \(c_2\). First, find the derivative:- \(y' = c_2 \sqrt{3} \cos(\sqrt{3}x) + 2\)- Substitute \(x = 1\) into both \(y\) and \(y'\) and solve the resulting equation to find \(c_2\).
Characteristic Equation
The characteristic equation is a critical tool in finding the homogeneous solution to linear differential equations. It helps in determining the nature of the solutions, such as whether they are exponential, oscillatory, or a mix of both.For a differential equation like \(y'' + 3y = 0\), assume solutions of the form \(y = e^{rx}\). This assumption transforms our task into an algebraic one.Substitute \(y = e^{rx}\) and its derivatives into the homogeneous equation:- \(y' = re^{rx}\)- \(y'' = r^2e^{rx}\)Plugging these into the equation gives:\[r^2 e^{rx} + 3 e^{rx} = 0\] Cancel \(e^{rx}\) since it is never zero, leaving:\[r^2 + 3 = 0\]Solving this yields complex roots \(r = \pm i \sqrt{3}\), indicating solutions with sine and cosine components. Thus, the characteristic equation helps define the homogeneous part \(y_h = c_1 \cos(\sqrt{3}x) + c_2 \sin(\sqrt{3}x)\). This understanding is vital to formulating complete solutions that fit both the structure of the differential equation and applied conditions.

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Most popular questions from this chapter

Define \(y=u(x) e^{x}\) so $$y^{\prime}=u e^{x}+u^{\prime} e^{x}, \quad y^{\prime \prime}=u^{\prime \prime} e^{x}+2 u^{\prime} e^{x}+u e^{x}$$ and $$y^{\prime \prime}-3 y^{\prime}+2 y=e^{x} u^{\prime \prime}-e^{x} u^{\prime}=5 e^{3 x}$$ If \(w=u^{\prime}\) we obtain the linear first-order equation \(w^{\prime}-w=5 e^{2 x}\) which has the integrating factor \(e^{-\int d x}=e^{-x}\) Now $$\frac{d}{d x}\left[e^{-x} w\right]=5 e^{x} \quad \text { gives } \quad e^{-x} w=5 e^{x}+c_{1}$$ Therefore \(w=u^{\prime}=5 e^{2 x}+c_{1} e^{x}\) and \(u=\frac{5}{2} e^{2 x}+c_{1} e^{x}+c_{2} .\) The general solution is $$y=u e^{x}=\frac{5}{2} e^{3 x}+c_{1} e^{2 x}+c_{2} e^{x}$$

Let \(m\) be the mass and \(k_{1}\) and \(k_{2}\) the spring constants. Then \(k=4 k_{1} k_{2} /\left(k_{1}+k_{2}\right)\) is the effective spring constant of the system. since the initial mass stretches one spring \(\frac{1}{3}\) foot and another spring \(\frac{1}{2}\) foot, using \(F=k s,\) we have \(\frac{1}{3} k_{1}=\frac{1}{2} k_{2}\) or \(2 k_{1}=3 k_{2}\). The given period of the combined system is \(2 \pi / \omega=\pi / 15,\) so \(\omega=30 .\) since a mass weighing 8 pounds is \(\frac{1}{4}\) slug, we have from \(w^{2}=k / m\) \\[30^{2}=\frac{k}{1 / 4}=4 k \quad \text { or } \quad k=225\\] We now have the system of equations \\[\begin{aligned}\frac{4 k_{1} k_{2}}{k_{1}+k_{2}} &=225 \\\2 k_{1} &=3 k_{2}\end{aligned}\\] Solving the second equation for \(k_{1}\) and substituting in the first equation, we obtain \\[\frac{4\left(3 k_{2} / 2\right) k_{2}}{3 k_{2} / 2+k_{2}}=\frac{12 k_{2}^{2}}{5 k_{2}}=\frac{12 k_{2}}{5}=225\\] Thus, \(k_{2}=375 / 4\) and \(k_{1}=1125 / 8 .\) Finally, the weight of the first mass is \\[32 m=\frac{k_{1}}{3}=\frac{1125 / 8}{3}=\frac{375}{8} \approx 46.88 \mathrm{lb}\\].

When the circuit is in resonance the form of \(q_{p}(t)\) is \(q_{p}(t)=A t \cos k t+B t \sin k t\) where \(k=1 / \sqrt{L C} .\) Substituting \(q_{p}(t)\) into the differential equation we find \\[q_{p}^{\prime \prime}+k^{2} q_{p}=-2 k A \sin k t+2 k B \cos k t=\frac{E_{0}}{L} \cos k t\\] Equating coefficients we obtain \(A=0\) and \(B=E_{0} / 2 k L .\) The charge is \\[q(t)=c_{1} \cos k t+c_{2} \sin k t+\frac{E_{0}}{2 k L} t \sin k t\\] The initial conditions \(q(0)=q_{0}\) and \(q^{\prime}(0)=i_{0}\) imply \(c_{1}=q_{0}\) and \(c_{2}=i_{0} / k .\) The current is \\[\begin{aligned}i(t) &=-c_{1} k \sin k t+c_{2} k \cos k t+\frac{E_{0}}{2 k L}(k t \cos k t+\sin k t) \\ &=\left(\frac{E_{0}}{2 k L}-q_{0} k\right) \sin k t+i_{0} \cos k t+\frac{E_{0}}{2 L} t \cos k t\end{aligned}\\]

The general solution of the differential equation is \\[ y=c_{1} \cos \sqrt{\frac{P}{E I}} x+c_{2} \sin \sqrt{\frac{P}{E I}} x+\frac{w_{0}}{2 P} x^{2}+\frac{w_{0} E I}{P^{2}} \\] Setting \(y(0)=0\) we obtain \(c_{1}=-w_{0} E I / P^{2},\) so that \\[ y=-\frac{w_{0} E I}{P^{2}} \cos \sqrt{\frac{P}{E I}} x+c_{2} \sin \sqrt{\frac{P}{E I}} x+\frac{w_{0}}{2 P} x^{2}+\frac{w_{0} E I}{P^{2}} \\] Setting \(y^{\prime}(L)=0\) we find \\[ c_{2}=\left(-\sqrt{\frac{P}{E I}} \frac{w_{0} E I}{P^{2}} \sin \sqrt{\frac{P}{E I}} L-\frac{w_{0} L}{P}\right) / \sqrt{\frac{P}{E I}} \cos \sqrt{\frac{P}{E I}} L \\]

From \(D x+D^{2} y=e^{3 t}\) and \((D+1) x+(D-1) y=4 e^{3 t}\) we obtain \(D\left(D^{2}+1\right) x=34 e^{3 t}\) and \(D\left(D^{2}+1\right) y=-8 e^{3 t}\) The solution is $$\begin{array}{l} y=c_{1}+c_{2} \sin t+c_{3} \cos t-\frac{4}{15} e^{3 t} \\ x=c_{4}+c_{5} \sin t+c_{6} \cos t+\frac{17}{15} e^{3 t} \end{array}.$$ Substituting into \((D+1) x+(D-1) y=4 e^{3 t}\) gives $$\left(c_{4}-c_{1}\right)+\left(c_{5}-c_{6}-c_{3}-c_{2}\right) \sin t+\left(c_{6}+c_{5}+c_{2}-c_{3}\right) \cos t=0$$ so that \(c_{4}=c_{1}, c_{5}=c_{3}, c_{6}=-c_{2},\) and $$x=c_{1}-c_{2} \cos t+c_{3} \sin t+\frac{17}{15} e^{3 t}.$$
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