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Chapter 3: Problem 33

Equations with the Dependent Variable Missing. For a scond order differential equation of the form \(y^{\prime \prime}=f\left(t, y^{\prime}\right),\) the substitution \(v=y^{\prime}, v^{\prime}=y^{\prime \prime}\) leads to a first order equation of the form \(v^{\prime}=f(t, v)\). If this equation can be solved for \(v\), then \(y\) can be obtained by integrating \(d y / d t=v .\) Note that one arbitrary constant is obtained in solving the first order equation for \(v,\) and a second is introduced in the integration for \(y\). In each of Problems 28 through 33 use this substitution to solve the given equation. $$ t^{2} y^{\prime \prime}=\left(y^{\prime}\right)^{2}, \quad t>0 $$

Short Answer

Expert verified
Question: Solve the second-order differential equation \(t^2y^{\prime \prime} = (y^\prime)^2\) for \(y(t)\) using the substitution \(v = y^\prime\) and \(v^\prime = y^{\prime \prime}\). Answer: The general solution to the given differential equation is \(y(t) = \frac{1}{C_1}(tC_1 - 1 + \ln |tC_1 - 1|) + C_2\), where \(C_1\) and \(C_2\) are arbitrary constants.

Step by step solution

01

Substitute \(v = y^\prime\) and \(v^\prime = y^{\prime \prime}\)

Replace \(y^{\prime}\) with \(v\) and \(y^{\prime \prime}\) with \(v^\prime\) in the given equation: $$ t^2v^\prime = v^2 $$
02

Solve first-order differential equation for \(v(t)\)

Rewrite the given equation and separate variables: $$ \frac{dv}{dt} = \frac{v^2}{t^2} $$ Now separate the variables and integrate both sides: $$ \int \frac{1}{v^2} dv = \int \frac{1}{t^2} dt $$ $$ -\frac{1}{v} = -\frac{1}{t} + C_1 $$ Solve for \(v\): $$ v = \frac{t}{tC_1 - 1} $$
03

Integrate \(v(t)\) to find \(y(t)\)

Integrate \(v(t)\) with respect to \(t\) to find \(y(t)\): $$ y = \int v dt = \int \frac{t}{tC_1 - 1} dt $$ Let \(u = tC_1 - 1 \Rightarrow du = C_1 dt \Rightarrow \frac{du}{C_1} = dt\). Our integral becomes: $$ y = \frac{1}{C_1} \int \frac{u + 1}{u} du $$ Now, integrate: $$ y = \frac{1}{C_1} (u + \ln |u|) + C_2 $$ Substitute back for \(u\): $$ y = \frac{1}{C_1}(tC_1 - 1 + \ln |tC_1 - 1|) + C_2 $$ This is the general solution to the given differential equation.

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Key Concepts

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

Second Order Differential Equation
A second order differential equation involves a second derivative of the unknown function. It typically takes the form \(y'' = f(t, y, y')\). In the original exercise, the equation \(t^2 y'' = (y')^2\) represents a special case where the dependent variable \(y\) is missing from \(f\), simplifying the equation.
The main challenge with second order differential equations is their complexity, as they involve higher derivatives that might not be easy to solve directly. Therefore, various strategies such as substitutions or transformations are used to simplify them into more manageable forms. Once transformed, these equations can sometimes be reduced to first order equations, which are generally easier to solve.
First Order Differential Equation
A first order differential equation involves only the first derivative of the unknown function. In our exercise, after using the substitution \(v = y'\) and \(v' = y''\), the second order equation simplifies into a first order differential equation: \(v' = \frac{v^2}{t^2}\).
First order equations like this often allow for methods such as separation of variables. These techniques transform the equation into a form that can be readily integrated. In our case, separating variables makes it possible to integrate and find \(v(t)\), which represents the rate of change of \(y\). First order equations are foundational since they frequently appear in more complex equations once higher derivatives are removed.
Integration
Integration is the process of finding a function with its derivative known, typically referred to as the antiderivative. In our problem, once we have \(v(t)\), integration allows us to find \(y(t)\).
We solve for \(v\) by integrating \(\int \frac{1}{v^2} dv\) and \(\int \frac{1}{t^2} dt\), leading to the relation \(-\frac{1}{v} = -\frac{1}{t} + C_1\). This step is essential as it finds a function \(v(t)\) that satisfies the equation. Once \(v(t)\) is known, the next step is to integrate it with respect to \(t\), thus obtaining \(y(t)\), the original dependent variable function.
  • Integration transforms the derivative function into the original function.
  • It introduces arbitrary constants, representing the general solution's family.
Substitution Method
The substitution method simplifies differential equations by introducing a new variable. In our equation, setting \(v = y'\) reduces the complexity from a second to a first order differential equation.
This transformation takes advantage of the structure of the equation where the dependent variable is missing. Rewriting the equation facilitates easier integration by separating variables. After substitution, adjusting the equation allows the use of simpler integration methods, resulting in an expression where \(v(t)\) and eventually \(y(t)\) are solvable.
  • By introducing new variables, harder problems become manageable.
  • It streamlines the process to find exact solutions by reducing the complexity of derivatives.

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Most popular questions from this chapter

Consider a vibrating system described by the initial value problem $$ u^{\prime \prime}+\frac{1}{4} u^{\prime}+2 u=2 \cos \omega t, \quad u(0)=0, \quad u^{\prime}(0)=2 $$ (a) Determine the steady-state part of the solution of this problem. (b) Find the amplitude \(A\) of the steady-state solution in terms of \(\omega\). (c) Plot \(A\) versus \(\omega\). (d) Find the maximum value of \(A\) and the frequency \(\omega\) for which it occurs.

By choosing the lower limitofation in Eq. ( 28 ) inthe textas the initial point \(t_{0}\), show that \(Y(t)\) becomes $$ Y(t)=\int_{t_{0}}^{t} \frac{y_{1}(s) y_{2}(t)-y_{1}(s) y_{2}(s)}{y_{1}(s) y_{2}^{2}(s)-y_{1}^{\prime}(s) y_{2}(s)} g(s) d s $$ Show that \(Y(t)\) is asolution of the initial value problem $$ L[y], \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ Thus \(Y\) can be identific d writh \(v\) in Problem \(21 .\)

In the spring-mass system of Problem \(31,\) suppose that the spring force is not given by Hooke's law but instead satisfies the relation $$ F_{s}=-\left(k u+\epsilon u^{3}\right) $$ where \(k>0\) and \(\epsilon\) is small but may be of either sign. The spring is called a hardening spring if \(\epsilon>0\) and a softening spring if \(\epsilon<0 .\) Why are these terms appropriate? (a) Show that the displacement \(u(t)\) of the mass from its equilibrium position satisfies the differential equation $$ m u^{\prime \prime}+\gamma u^{\prime}+k u+\epsilon u^{3}=0 $$ Suppose that the initial conditions are $$ u(0)=0, \quad u^{\prime}(0)=1 $$ In the remainder of this problem assume that \(m=1, k=1,\) and \(\gamma=0\). (b) Find \(u(t)\) when \(\epsilon=0\) and also determine the amplitude and period of the motion. (c) Let \(\epsilon=0.1 .\) Plot (a numerical approximation to) the solution. Does the motion appear to be periodic? Estimate the amplitude and period. (d) Repeat part (c) for \(\epsilon=0.2\) and \(\epsilon=0.3\) (e) Plot your estimated values of the amplitude \(A\) and the period \(T\) versus \(\epsilon\). Describe the way in which \(A\) and \(T\), respectively, depend on \(\epsilon\). (f) Repeat parts (c), (d), and (e) for negative values of \(\epsilon .\)

Use the substitution introduced in Problem 38 in Section 3.4 to solve each of the equations \(t^{2} y^{\prime \prime}-3 t y^{\prime}+4 y=0, \quad t>0\)

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}-(x-0.1875) y=0, \quad x>0 ; \quad y_{1}(x)=x^{1 / 4} e^{2 \sqrt{x}}\)
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