91Ó°ÊÓ

Let \(u=y^{\prime}\) so that \(u^{\prime}=y^{\prime \prime} .\) The equation becomes \(\left(u^{\prime}\right)^{2}+u^{2}=1\) which results in \(u^{\prime}=\pm \sqrt{1-u^{2}}\). To solve \(u^{\prime}=\sqrt{1-u^{2}}\) we separate variables: \\[ \begin{aligned} \frac{d u}{\sqrt{1-u^{2}}}=d x & \Longrightarrow \sin ^{-1} u=x+c_{1} \Longrightarrow u=\sin \left(x+c_{1}\right) \\ & \Longrightarrow y^{\prime}=\sin \left(x+c_{1}\right) \end{aligned} \\] When \(x=\pi / 2, y^{\prime}=\sqrt{3} / 2,\) so \(\sqrt{3} / 2=\sin \left(\pi / 2+c_{1}\right)\) and \(c_{1}=-\pi / 6 .\) Thus \\[ y^{\prime}=\sin \left(x-\frac{\pi}{6}\right) \Rightarrow y=-\cos \left(x-\frac{\pi}{6}\right)+c_{2} \\] When \(x=\pi / 2, y=1 / 2,\) so \(1 / 2=-\cos (\pi / 2-\pi / 6)+c_{2}=-1 / 2+c_{2}\) and \(c_{2}=1 .\) The solution of the initial-value problem is \(y=1-\cos (x-\pi / 6)\) To solve \(u^{\prime}=-\sqrt{1-u^{2}}\) we separate variables: \\[ \begin{aligned} \frac{d u}{\sqrt{1-u^{2}}}=-d x & \Longrightarrow \cos ^{-1} u=x+c_{1} \\ & \Longrightarrow u=\cos \left(x+c_{1}\right) \Longrightarrow y^{\prime}=\cos \left(x+c_{1}\right) \end{aligned} \\] When \(x=\pi / 2, y^{\prime}=\sqrt{3} / 2,\) so \(\sqrt{3} / 2=\cos \left(\pi / 2+c_{1}\right)\) and \(c_{1}=-\pi / 3 .\) Thus \\[ y^{\prime}=\cos \left(x-\frac{\pi}{3}\right) \Rightarrow y=\sin \left(x-\frac{\pi}{3}\right)+c_{2} \\] When \(x=\pi / 2, y=1 / 2,\) so \(1 / 2=\sin (\pi / 2-\pi / 3)+c_{2}=1 / 2+c_{2}\) and \(c_{2}=0 .\) The solution of the initial-value problem is \(y=\sin (x-\pi / 3).\)

Step by step solution

01

Replace Derivatives

We start by setting the substitution: let \( u = y' \). This substitution implies that \( u' = y'' \). The original differential equation given as \((u')^2 + u^2 = 1\) becomes a simpler form involving \( u \) and its derivative \( u' \).

02

Solve for u' Considering Positive Case

Consider the case \( u' = \sqrt{1-u^2} \). Separate variables to integrate: \( \frac{du}{\sqrt{1-u^2}} = dx \). This leads us to the integral which results in \( \sin^{-1}(u) = x + c_1 \). Solving for \( u \), we find \( u = \sin(x + c_1) \).

03

Solve for y' Given Initial Condition

At \( x = \frac{\pi}{2} \), \( y' = u = \frac{\sqrt{3}}{2} \). Thus, \( \sin(\frac{\pi}{2} + c_1) = \frac{\sqrt{3}}{2} \), giving us \( c_1 = -\frac{\pi}{6} \). Therefore, \( y' = \sin(x - \frac{\pi}{6}) \).

04

Integrate to Find y

Integrate \( y' = \sin(x - \frac{\pi}{6}) \) to find \( y \). The integration result is \( y = -\cos(x - \frac{\pi}{6}) + c_2 \).

05

Determine Constant c_2

Given \( x = \frac{\pi}{2} \), \( y = \frac{1}{2} \). So we solve \( \frac{1}{2} = -\cos(\frac{\pi}{2} - \frac{\pi}{6}) + c_2 \), leading to \( c_2 = 1 \). Therefore, the solution is \( y = 1 - \cos(x - \frac{\pi}{6}) \).

06

Solve for u' Considering Negative Case

Now consider the case \( u' = -\sqrt{1-u^2} \). Separate variables: \( \frac{du}{\sqrt{1-u^2}} = -dx \). This results in \( \cos^{-1}(u) = x + c_1 \), or \( u = \cos(x + c_1) \).

07

Solve for y' in Negative Case

With \( x = \frac{\pi}{2} \) and \( y' = \frac{\sqrt{3}}{2} \), we get \( \cos(\frac{\pi}{2} + c_1) = \frac{\sqrt{3}}{2} \) and solve to find \( c_1 = -\frac{\pi}{3} \). Thus, \( y' = \cos(x - \frac{\pi}{3}) \).

08

Integrate Negative Case to Find y

Integrate \( y' = \cos(x - \frac{\pi}{3}) \) to derive \( y = \sin(x - \frac{\pi}{3}) + c_2 \).

09

Determine New Constant c_2

Using the initial value \( x = \frac{\pi}{2} \), \( y = \frac{1}{2} \), solve \( \frac{1}{2} = \sin(\frac{\pi}{2} - \frac{\pi}{3}) + c_2 \), which results in \( c_2 = 0 \). Therefore, \( y = \sin(x - \frac{\pi}{3}) \).

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Key Concepts

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

Initial Value Problem

In differential equations, an initial value problem (IVP) involves finding a function that satisfies the differential equation and meets specified conditions at given points. These conditions are known as initial values. This type of problem is commonly structured to include not only the differential equation but also one or more initial conditions, typically:

• A value for the function at a certain point.
• A value for the function's derivatives at certain points.

These initial conditions are crucial because they allow us to find the specific solution among the infinite set of possible solutions that satisfy the differential equation. For instance, consider a differential equation involving the motion of an object under the influence of gravity. We would need initial conditions like the height and velocity of the object at a starting time to determine its precise behavior.
In the given exercise, the IVP specified that at \(x = \frac{\pi}{2}\), the derivative y' should be \(\frac{\sqrt{3}}{2}\) and y should be \(\frac{1}{2}\). These conditions were used at various stages of the solution to precisely determine the constants for our solution, leading to a unique answer.

Separation of Variables

Separation of variables is a common technique used to solve ordinary differential equations (ODEs). The primary goal is to manipulate the equation so that all terms involving the dependent variable appear on one side, while all terms involving the independent variable appear on the other. Here’s a simple way to understand it:

• If the ODE is of the form \( \frac{dY}{dx} = g(x)h(Y) \), we aim to rewrite it as \( \frac{dY}{h(Y)} = g(x) \, dx \).
• Once separated, integrate both sides independently.

This method works well for ODEs that can be factored into separate functions of each variable. It's particularly useful in cases where the equation describes physical phenomena with distinct rates of change.
In the context of this exercise, the differential equation \( u' = \sqrt{1-u^2} \) was solved by moving terms related to u to one side and those related to x to the other. This resulted in integrating both sides to find a solution involving inverse trigonometric functions. Through integration, we arrived at solutions that explicitly depend on x, aiding the resolution of the initial value problem.

Trigonometric Integration

Trigonometric integration is an essential technique used to solve integrals of expressions involving trigonometric functions. It involves using identities and substitutions to simplify integrals into forms that are more manageable. Here are some key points:

• Trigonometric identities, like \( \sin^2 x + \cos^2 x = 1 \), help reduce the complexity of integrals.
• Substitutions are often used, especially the inverse trigonometric functions, such as \( \sin^{-1}(x) \), \( \cos^{-1}(x) \), and \( \tan^{-1}(x) \).

These methods transform difficult integrations into solvable ones by expressing them in terms of standard integral forms.
In the problem at hand, the separation of variables process led to integrating trigonometric expressions. For example, integrating \( \frac{du}{\sqrt{1-u^2}} \) results in an inverse sine function \( \sin^{-1}(u) \), which directly contributed to finding the function that satisfies the given initial value problem.