Question

In each of the following cases, show that the given function \(Y(x)\) satisfies the associated differential equation. Then determine the value of \(c\) required by the initial condition. Finally, with reference to the general format in (8.7), identify \(f(x, z)\) for each differential equation. (a) \(\quad Y^{\prime}(x)=-Y(x)+\sin (x)+\cos (x), \quad Y(0)=1 ; \quad Y(x)=\sin (x)+c e^{-x}\) (b) \(Y^{\prime}(x)=\left(Y(x)-Y(x)^{2}\right) / x, \quad Y(1)=2 ; \quad Y(x)=x /(x+c), \quad x>0\) (c) \(Y^{\prime}(x)=\cos ^{2}(Y(x)), \quad Y(0)=\pi / 4 ; \quad Y(x)=\tan ^{-1}(x+c)\) (d) \(\quad Y^{\prime}(x)=Y(x)[Y(x)-1], \quad Y(0)=\frac{1}{2} ; \quad Y(x)=1 /\left(1+c e^{x}\right)\)

Short answer

The functions all satisfy their differential equations. The constants are: (a) c = 1; (b) c = -1/2; (c) c = 1; (d) c = 1. Functions f(x, z) are identified as required.

Step-by-step solution

Verification of Differential Equation (a)

Differentiate the given function: \[Y(x) = \sin(x) + c e^{-x}\] \[Y'(x) = \cos(x) - c e^{-x}\]Substitute into the differential equation: \[Y'(x) = -Y(x) + \sin(x) + \cos(x)\] \[\cos(x) - c e^{-x} = -\left( \sin(x) + c e^{-x} \right) + \sin(x) + \cos(x)\]\[\cos(x) - c e^{-x} = - \sin(x) - c e^{-x} + \sin(x) + \cos(x)\]It matches, so the function satisfies the differential equation.

Determine Constant c for (a)

Apply the initial condition: \[Y(0) = 1\] Using the function: \[1 = \sin(0) + c e^{0}\]\[1 = 0 + c\]\[c = 1\]

Function Identification for (a)

Refer to format (8.7): \[Y'(x) = f(x, Y(x))\] \[f(x, z) = -z + \sin(x) + \cos(x)\]

Verification of Differential Equation (b)

Differentiate the given function: \[Y(x) = \frac{x}{x+c}\]\[Y'(x) = \frac{(x+c) - x}{(x+c)^{2}} = \frac{c}{(x+c)^{2}}\]Substitute into the differential equation:\[Y'(x) = \frac{Y(x) - Y(x)^{2}}{x}\] \[\frac{c}{(x+c)^{2}} = \frac{\frac{x}{x+c} - \left( \frac{x}{x+c} \right)^2}{x}\]It simplifies correctly, so the function satisfies the differential equation.

Determine Constant c for (b)

Apply the initial condition: \[Y(1) = 2\] Using the function: \[2 = \frac{1}{1+c}\]\[2(1+c) = 1\]\[c = -\frac{1}{2}\]

Function Identification for (b)

Refer to format (8.7): \[Y'(x) = f(x, Y(x))\] \[f(x, z) = \frac{z - z^{2}}{x}\]

Verification of Differential Equation (c)

Differentiate the given function: \[Y(x) = \tan^{-1}(x+c)\]Use the derivative of arctangent: \[Y'(x) = \frac{1}{1 + (x+c)^2}\]Substitute into the differential equation: \[Y'(x) = \cos^{2}(Y(x))\] Knowing \(\cos(\tan^{-1}(u)) = \frac{1}{\sqrt{1+u^2}}\), then \(\cos^{2}(\tan^{-1}(u)) = \frac{1}{1+u^2}\)\[Y'(x) = \frac{1}{1 + (x+c)^2}\]It matches, so the function satisfies the differential equation.

Determine Constant c for (c)

Apply the initial condition:\[Y(0) = \frac{\pi}{4}\] Using the function: \[\frac{\pi}{4} = \tan^{-1}(0+c)\]\[\frac{\pi}{4} = \tan^{-1}(c)\]It's known: \(\tan(\frac{\pi}{4}) = 1\), so \(c = 1\)

Function Identification for (c)

Refer to format (8.7): \[Y'(x) = f(x, Y(x))\] \[f(x, z) = \cos^{2}(z)\]

Verification of Differential Equation (d)

Differentiate the given function: \[Y(x) = \frac{1}{1 + c e^x}\]Using the quotient rule: \[Y'(x) = \frac{-c e^x}{(1 + c e^x)^2}\]Substitute into the differential equation: \[Y'(x) = Y(x) (Y(x) - 1)\] \[\frac{-c e^x}{(1 + c e^x)^2} = \frac{1}{1 + c e^x} \left( \frac{1}{1 + c e^x} - 1 \right)\]It matches, so the function satisfies the differential equation.

Determine Constant c for (d)

Apply the initial condition:\[Y(0) = \frac{1}{2}\] Using the function: \[\frac{1}{2} = \frac{1}{1 + c}\]\[2 = 1 + c\]\[c = 1\]

Function Identification for (d)

Refer to format (8.7): \[Y'(x) = f(x, Y(x))\] \[f(x, z) = z (z-1)\]

Key concepts

Differential Equations

A differential equation is a mathematical equation that involves derivatives of a function. These equations help in describing the rate at which things change. For example, in physics, they can describe how the position of a moving object changes over time. Differential equations can be first-order, second-order, and so on, depending on the highest derivative involved.
To solve a differential equation, you often need to find a function (or a set of functions) that satisfies the equation. This involves steps like differentiating and substituting back to check if the equation holds. This process is key in determining if a given solution is correct.

Initial Conditions

Initial conditions are extra pieces of information provided alongside a differential equation, usually in the form of specific values for the function and its derivatives at given points. These conditions help pinpoint the exact solution out of possibly many. Think of them as additional clues that narrow down the possibilities.
For example, consider the initial condition given as \(Y(0) = 1\). This means that when \(x = 0\), the value of the function \(Y(x)\) should be 1. When solving, you use this information to find any unknown constants. Initial conditions make sure the solution fits the specified scenario.

Function Identification

Function identification involves recognizing the given function in the format of the differential equation provided. The general format is often given as \(Y'(x) = f(x, Y(x))\). The function \(f(x, z)\) expresses how the derivative of \(Y\) is related to \(x\) and \(Y\) itself.
This step helps to understand the nature of the differential equation and lays the groundwork for further calculations. It involves careful analysis to see how the given function matches the form of \(f(x, z)\). This identification is crucial for verifying and solving the equation effectively.

Quotient Rule

The quotient rule is a method used in calculus to find the derivative of a function that is the ratio of two differentiable functions. It states that if you have a function \(Y(x) = \frac{u(x)}{v(x)}\), the derivative, \(Y'(x)\), is:
\[Y'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}\]
This rule is handy when dealing with fractional forms in differential equations. For instance, if you have \(Y(x) = \frac{1}{1 + c e^x}\), you use the quotient rule to find its derivative. It involves differentiating both the numerator and the denominator separately and then combining them according to the rule. Understanding this rule simplifies solving several types of differential equations.