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In this problem we determine conditions on \(p\) and \(q\) such that \(\mathrm{Eq}\). (i) can be transformed into an equation with constant coefficients by a change of the independent variable, Let \(x=u(t)\) be the new independent variable, with the relation between \(x\) and \(t\) to be specified later. (a) Show that $$ \frac{d y}{d t}=\frac{d x}{d t} \frac{d y}{d x}, \quad \frac{d^{2} y}{d t^{2}}=\left(\frac{d x}{d t}\right)^{2} \frac{d^{2} y}{d x^{2}}+\frac{d^{2} x}{d t^{2}} \frac{d y}{d x} $$ (b) Show that the differential equation (i) becomes $$ \left(\frac{d x}{d t}\right)^{2} \frac{d^{2} y}{d x^{2}}+\left(\frac{d^{2} x}{d t^{2}}+p(t) \frac{d x}{d t}\right) \frac{d y}{d x}+q(t) y=0 $$ (c) In order for Eq. (ii) to have constant coefficients, the coefficients of \(d^{2} y / d x^{2}\) and of \(y\) must be proportional. If \(q(t)>0,\) then we can choose the constant of proportionality to be \(1 ;\) hence $$ x=u(t)=\int[q(t)]^{1 / 2} d t $$ (d) With \(x\) chosen as in part (c) show that the coefficient of \(d y / d x\) in Eq. (ii) is also a constant, provided that the expression $$ \frac{q^{\prime}(t)+2 p(t) q(t)}{2[q(t)]^{3 / 2}} $$ $$ \frac{q^{\prime}(t)+2 p(t) q(t)}{2[q(t)]^{3 / 2}} $$ is a constant. Thus Eq. (i) can be transformed into an equation with constant coefficients by a change of the independent variable, provided that the function \(\left(q^{\prime}+2 p q\right) / q^{3 / 2}\) is a constant. How must this result be modified if \(q(t)<0 ?\)

Short Answer

Expert verified
Answer: The differential equation can be transformed into one with constant coefficients provided that the expression (q'(t) + 2p(t)q(t)) / [q(t)]^{3/2} is a constant.

Step by step solution

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01

Derive expressions for dy/dt and d²y/dt²

Differentiate y with respect to t using the chain rule: $$\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}$$ Now differentiate y once more with respect to t: $$\frac{d^{2}y}{dt^{2}} = \frac{d}{dt} \left(\frac{dy}{dx} \frac{dx}{dt}\right) = \left(\frac{dx}{dt}\right)^{2} \frac{d^{2}y}{dx^{2}} + \frac{d^{2}x}{dt^{2}} \frac{dy}{dx}$$
02

Substitute the expressions into the differential equation

Substitute the expressions for dy/dt and d²y/dt² into the given differential equation: $$\left(\frac{dx}{dt}\right)^{2} \frac{d^{2}y}{dx^{2}} + \left(\frac{d^{2}x}{dt^{2}} + p(t) \frac{dx}{dt}\right) \frac{dy}{dx} + q(t)y = 0$$
03

Determine the conditions for proportional coefficients

We want the coefficients of \(d^{2}y/dx^{2}\) and y to be proportional in order for the equation to have constant coefficients. If \(q(t) > 0\), we can choose the constant of proportionality to be 1: $$x = u(t) = \int[q(t)]^{1/2} dt$$
04

Show that the coefficient of dy/dx is constant

Under the conditions found in Step 3, we want to show that the coefficient of \(dy/dx\) in the equation is also a constant. We are given the expression $$\frac{q'(t) + 2p(t)q(t)}{2[q(t)]^{3/2}}$$ In order for the equation to have constant coefficients, we require this expression to be a constant. Hence, the differential equation can be transformed into one with constant coefficients via a change of independent variable, provided that \(\left(q'+2pq\right)/q^{3/2}\) is a constant.
05

Handle the case when q(t) < 0

If \(q(t) < 0\), then the given transformed equation will have imaginary or complex coefficients. To avoid this situation, we could change the sign of q(t) and proceed with the same method. In this case, the conditions on p(t) and q(t) will still hold, but the equation will have constant, real, and negative coefficients.

Key Concepts

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

Change of Variables in Differential Equations
In solving differential equations, one powerful technique involves changing the independent variable to simplify the equation. This process is known as a change of variables. It can help transform complex equations into forms that are easier to solve, such as those with constant coefficients.

For instance, consider a differential equation that's difficult to tackle in its current form. By cleverly choosing a new independent variable, say, changing from variable \( t \) to \( x \), where \( x \) is a function of \( t \), \( x = u(t) \), we can reformulate the equation. It requires deriving the relationships between derivatives with respect to \( t \) and \( x \), which is usually done using the chain rule of calculus.

The chain rule states that the derivative of a function with respect to one variable (like \( t \) in our example) can be expressed as the product of the derivative with respect to another variable (like \( x \) here) and the derivative of that second variable with respect to the first. This concept allows us to re-express derivatives, and consequently, the differential equation in terms of the new variable, potentially simplifying it.
Constant Coefficient Differential Equation
A constant coefficient differential equation is one where the coefficients of the terms involving derivatives are constants, rather than functions of the independent variable. These types of equations are desirable due to their relatively straightforward analytic and numerical solution methods.

To illustrate, if we have a second-order differential equation with variable coefficients \( p(t) \) and \( q(t) \) and we want it to have constant coefficients, we may aim to manipulate the equation until the coefficients of the derivative terms do not vary with the independent variable. In the exercise at hand, the goal is to find a transformation \( u(t) \) such that when we adopt this new variable as the independent variable, the resulting differential equation has constant coefficients. This process involves significant insights into the original equation's structure and often requires identification of an appropriate substitution that normalizes or simplifies the variable coefficients.
Proportional Coefficients for Constant Terms
When we adjust differential equations through a change of variable, our aim is often to have the coefficients of both the highest derivative term and the zeroth derivative term (the term with just the dependent variable, no derivatives) become constants. Moreover, for the differential equation to have constant coefficients after transformation, these constants should be proportional.

By setting a proportionality condition, as seen in the exercise, where the coefficient of \( d^2y/dx^2 \) and the term with \( y \) are both 1, we add constraints to our problem. This proportionality leads us to determine the function \( u(t) \) that will accomplish our goal. As indicated in the solution, if \( q(t) > 0 \) and by selecting \( x = \( \int[q(t)]^{1/2} dt \) \) we get constant coefficients for both terms, which simplifies the equation substantially.

The proportionality of coefficients is crucial because it allows for a standard form of the differential equation, enabling the use of general solutions that have been widely studied. If the coefficients are not proportional or constant, the equation could still be solvable, but it would require different methods that can handle variable coefficients, and these methods can be more complex and less intuitive.

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Most popular questions from this chapter

A series circuit has a capacitor of \(10^{-5}\) farad, a resistor of \(3 \times 10^{2}\) ohms, and an inductor of 0.2 henry. The initial charge on the capacitor is \(10^{-6}\) coulomb and there is no initial current. Find the charge \(Q\) on the capacitor at any time \(t .\)

A spring-mass system with a hardening spring (Problem 32 of Section 3.8 ) is acted on by a periodic external force. In the absence of damping, suppose that the displacement of the mass satisfies the initial value problem $$ u^{\prime \prime}+u+\frac{1}{5} u^{3}=\cos \omega t, \quad u(0)=0, \quad u^{\prime}(0)=0 $$ (a) Let \(\omega=1\) and plot a computer-generated solution of the given problem. Does the system exhibit a beat? (b) Plot the solution for several values of \(\omega\) between \(1 / 2\) and \(2 .\) Describe how the solution changes as \(\omega\) increases.

Determine the general solution of $$ y^{\prime \prime}+\lambda^{2} y=\sum_{m=1}^{N} a_{m} \sin m \pi t $$ $$ \text { where } \lambda>0 \text { and } \lambda \neq m \pi \text { for } m=1, \ldots, N $$

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ \begin{array}{l}{x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=g(x), \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x, \quad y_{2}(x)=} \\\ {x^{-1 / 2} \cos x}\end{array} $$

Use the method of Problem 32 to solve the given differential $$ y^{\prime \prime}+2 y^{\prime}=3+4 \sin 2 t \quad \text { (see Problem } 4 \text { ) } $$

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