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When you finish this section, you should be able to:
An ordinary differential equation is an equation involving an independent variable \(x\), a dependent variable \(y\), and derivatives of \(y\) with respect to \(x.\) Examples of ordinary differential equations are:
Other variables besides \(x\) and \(y\) are often used. If \(s=f(t)\), then \(\dfrac{d^2s}{dt^2}+t=0\) is a second-order differential equation.
The order of a differential equation is the order of the highest-order derivative of \(y\) appearing in the equation. For the equations above, (a), (b), and (f) are first-order, (c) and (e) are second-order, and (d) is a third-order differential equation. The exponent of the highest-order derivative in a differential equation is called the degree of the equation. For example, equations (a)–(e) above are of degree 1, and equation (f) is of degree 2. Being able to recognize the order and degree of a differential equation are important steps in finding its solution.
An important class of differential equations are linear differential equations. If \(P_{0},P_{1},\ldots ,P_{n}\) and \(Q\) are functions of the variable \(x\), then an equation of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{P_{0}( x) \dfrac{d^{n}y}{dx^{n}}+P_{1}(x)\dfrac{d^{n-1}y}{ dx^{n-1}}+\cdots +P_{n-1}(x)\dfrac{dy}{dx}+P_{n}(x)y=Q(x) }} \]
is a linear differential equation of order \(n\). The term linear is used to describe these differential equations because each derivative including \(y\) (which we treat as a derivative of order zero) is of degree 1, no term contains more than one derivative, and no term contains a product of \(y\) and any derivative of \(y\) up to the \(n\)th derivative. For example, \[ \begin{equation*} x^{2}\frac{d^{2}y}{dx^{2}}+8\frac{dy}{dx}-5x^{3}y=x+10 \end{equation*} \]
is a second-order linear differential equation.
What are the order and degree of the differential equation \(xe^{x}\dfrac{d^{2}y}{dx^{2}}-5e^{x}\dfrac{dy}{dx}+4x^{3}y=0.\) Is it linear?
Solution This is a second-order differential equation of degree 1. Since \(P_{0}( x) =xe^{x},\) \(P_{1}( x) =-5e^{x}\), and \(P_{2}( x) =4x^{3}\) are all functions of \(x\) alone and no term contains more than one derivative, the differential equation is linear.
Problem 9.
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A function \(y=f(x)\) is called a solution to a differential equation on an interval \(I\) if, when \(y\) and its derivatives are substituted into the equation, the equation is satisfied for all \(x\) in the interval \(I.\)
The general solution of a differential equation represents all the solutions to the differential equation and is in the form of an equation with arbitrary constants. The number of arbitrary constants in the general solution agrees with the order of the differential equation.
As an example, consider the second-order differential equation \[ \begin{equation*} \frac{d^{2}y}{dx^{2}}+y=0 \end{equation*} \]
It can be shown that the general solution has the form \(y=C_{1}\sin x+C_{2}\cos x,\) where \(C_{1}\) and \(C_{2}\) are constants.
Any solution of a differential equation obtained from the general solution byeak assigning values to the arbitrary constants is called a particular solution. So, for this example, choosing \(C_{1}\,{=}\,6\) and \(C_{2}\,{=}\,9\) yields the particular solution \(y\,{=}\,6\sin x\,{+}\,9\cos x.\)
Solution (a) For \(y=f(x)=x^{4}-5x+1,\) we have \[ \begin{equation*} y^{\prime} =4x^{3}-5\qquad y^{\prime \prime} =12x^{2} \end{equation*} \]
The function \(f\) satisfies the differential equation \(y^{\prime \prime} -12x^{2}=0\) and so \(f\) is a solution.
(b) For \(y=g(x)=x^{4}+C_{1}x+C_{2}\), we have \[ \begin{equation*} y^{\prime} =4x^{3}+C_{1}\qquad y^{\prime \prime} =12x^{2} \end{equation*} \]
The function \(g\) satisfies the differential equation \(y^{\prime \prime} -12x^{2}=0\), so \(g\) is a solution.
Implicit differentiation is discussed in Section 3.2, pp. 209-212.
(c) Differentiate \(x^{2}-x^{3}y+3y^{4}=C\) implicitly with respect to \(x\) to find \(\dfrac{dy}{dx}\). \[ \begin{eqnarray*} 2x-x^{3}\dfrac{dy}{dx}-3x^{2}y+12y^{3}\dfrac{dy}{dx} &=&0 \\ ( 12y^{3}-x^{3}) \dfrac{dy}{dx} &=&3x^{2}y-2x \\[3pt] \dfrac{dy}{dx} &=&\dfrac{3x^{2}y-2x}{12y^{3}-x^{3}} \end{eqnarray*} \]
The function \(y=f( x) \) defined by the equation \(x^{2}-x^{3}y+3y^{4}=C\) satisfies the first-order differential equation and so is a solution.
Problems 21 and 23.
*It can be shown that this is the general solution of the differential equation.