/*! 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 12 In this case we have \(y(0)=c_{1... [FREE SOLUTION] | 91Ó°ÊÓ

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In this case we have \(y(0)=c_{1}=1, y^{\prime}(1)=2 c_{2}=6\) so that \(c_{1}=1\) and \(c_{2}=3 .\) The solution is \(y=1+3 x^{2}.\)

Short Answer

Expert verified
The solution to the problem is \( y = 1 + 3x^2 \).

Step by step solution

01

Identify Given Values

We are given initial conditions: \( y(0) = c_1 = 1 \) and \( y'(1) = 2c_2 = 6 \). From this, we know \( c_1 = 1 \) and need to find \( c_2 \).
02

Solve for \( c_2 \) using Given Condition

Given that \( y'(1) = 2c_2 = 6 \), substitute \( 1 \) into the derivative to find \( c_2 \). So, \( 2c_2 = 6 \rightarrow c_2 = \frac{6}{2} = 3 \).
03

Use Calculated Constants to Formulate a General Solution

The problem states that the solution is \( y = 1 + 3x^2 \). Given \( c_1 = 1 \) and \( c_2 = 3 \), this equation fits the form \( y = c_1 + c_2x^2 \), confirming the solution.

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Key Concepts

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

Initial Value Problems
Initial value problems are fundamental in the study of differential equations. They involve finding a solution to a differential equation that satisfies initial conditions. These conditions specify the values of the function and, in some cases, its derivatives at a particular point.
For example, if we have the differential equation expressed as a function, we might be given initial conditions such as the function value at a particular point and the derivative at another point.
This information allows us to tailor a general solution to fit specific requirements. When solving initial value problems, the process usually goes like this:
  • Identify the differential equation and the initial conditions.
  • Solve the differential equation to find the general solution.
  • Apply the initial conditions to determine the specific solution that fits the problem.
This process essentially ties a specific curve to a general family of solutions, making it unique and tailored to the problem at hand.
Second-Order Differential Equations
Second-order differential equations are equations involving the second derivative of a function. These types often appear in physics and engineering, where they model dynamic systems like vibrations, circuits, or motion.
The general form of a second-order differential equation is:\[ y'' + p(x) y' + q(x) y = g(x) \]Where:
  • \( y'' \) is the second derivative of the function \( y \).
  • \( p(x) \), \( q(x) \), and \( g(x) \) are functions of \( x \).
Solving these equations involves techniques like characteristic equations, particularly when the coefficients are constant. The complexity increases when coefficients vary with \( x \).
Simpler forms may be solved using classical methods or by transforming into a system of first-order equations.Substituting initial values into the general solution allows determination of any constant terms. This process ties into solving initial value problems.
Solutions of Differential Equations
Solutions to differential equations provide us with functions that satisfy the given equations and any initial conditions. They can be expressed in different forms depending on the equation's nature and complexity.
Some solutions are explicit, where the function is isolated on one side of the equation. Others may be implicit if directly solving for the function is complex. Finding a solution generally requires integration techniques or, sometimes, special functions when equations are non-standard. Here are key steps in finding solutions:
  • Determine the type of differential equation you are dealing with (e.g., linear, separable, exact).
  • Choose the appropriate method to find the solution (e.g., substitution, integrating factors, series solutions).
  • Apply any initial or boundary conditions to ensure the solution fits the requirements of the problem.
In our original exercise, solving the differential equation involved determining constants using initial conditions, resulting in an explicit solution formulation for specific parameters.

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Most popular questions from this chapter

If the constant \(-c_{1}^{2}\) is used instead of \(c_{1}^{2},\) then, using partial fractions \\[ y=-\int \frac{d x}{x^{2}-c_{1}^{2}}=-\frac{1}{2 c_{1}} \int\left(\frac{1}{x-c_{1}}-\frac{1}{x+c_{1}}\right) d x=\frac{1}{2 c_{1}} \ln \left|\frac{x+c_{1}}{x-c_{1}}\right|+c_{2} \\] Alternatively, the inverse hyperbolic tangent can be used.

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)\\].

(a) The auxiliary equation is \(m^{2}-64 / L=0\) which has roots \(\pm 8 / \sqrt{L}\). Thus, the general solution of the differential equation is \(x=c_{1} \cosh (8 t / \sqrt{L})+c_{2} \sinh (8 t / \sqrt{L})\) (b) Setting \(x(0)=x_{0}\) and \(x^{\prime}(0)=0\) we have \(c_{1}=x_{0}, 8 c_{2} / \sqrt{L}=0 .\) Solving for \(c_{1}\) and \(c_{2}\) we get \(c_{1}=x_{0}\) and \(c_{2}=0,\) so \(x(t)=x_{0} \cosh (8 t / \sqrt{L})\) (c) When \(L=20\) and \(x_{0}=1, x(t)=\cosh (4 t / \sqrt{5}) .\) The chain will last touch the peg when \(x(t)=10\) Solving \(x(t)=10\) for \(t\) we get \(t_{1}=\frac{1}{4} \sqrt{5} \cosh ^{-1} 10 \approx 1.67326 .\) The velocity of the chain at this instant is \(x^{\prime}\left(t_{1}\right)=12 \sqrt{11 / 5} \approx 17.7989 \mathrm{ft} / \mathrm{s}\).

The differential equation should have the form \(y^{\prime \prime}+k^{2} y=0\) where \(k=1\) so that the period of the solution is \(2 \pi .\) Thus, the differential equation is (d).

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 \(x(t)=\frac{\sqrt{3}}{12} \sin 8 \sqrt{3} t\).

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