Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Finding general solutions Find the general solution of each differential equation. Use \(C, C_{1}, C_{2}, \ldots\) to denote arbitrary constants. $$y^{\prime \prime}(x)=\frac{x}{\left(1-x^{2}\right)^{3 / 2}}$$

Short Answer

Expert verified
The general solution to the given differential equation is: $$y(x) = \arcsin(x) + C_1x + C_2$$

Step by step solution

01

Integrate the given equation once

Since we're given the second derivative, \(y^{\prime \prime}(x)\), let's integrate once with respect to \(x\) to find the first derivative, \(y'(x)\): $$y'(x) = \int y^{\prime \prime}(x) dx = \int \frac{x}{\left(1-x^{2}\right)^{3 / 2}} dx$$ This integral is non-trivial, so we will use a substitution method. Let \(u = 1-x^2\), then \(du = -2x dx\). We have: $$y'(x) = \int \frac{-1}{2} \cdot \frac{-2x dx}{u^{3/2}} = - \frac{1}{2} \int u^{-\frac{3}{2}} du$$ Now we can integrate \(u\): $$y'(x) = - \frac{1}{2}\left[- \frac{2}{\sqrt{u}}\right] + C_1 = \frac{1}{\sqrt{1-x^2}} + C_1$$ Here, we have integrated once and obtained \(y'(x)\).
02

Integrate again to find y(x)

We now need to find \(y(x)\) by integrating the expression for \(y'(x)\): $$y(x) = \int y'(x) dx = \int \left(\frac{1}{\sqrt{1-x^2}} + C_1\right) dx$$ We will split this integral into two parts and integrate each separately: $$y(x) = \int \frac{1}{\sqrt{1-x^2}} dx + C_1 \int dx$$ First, we can integrate the second part easily, and it results in \(C_1x\). Now we need to solve the first integral, which is a standard integral. Recall that \(\int \frac{1}{\sqrt{1-x^2}} dx = \arcsin(x)\). Thus, we have: $$y(x) = \arcsin(x) + C_1x + C_2$$ Here, we have found the general solution to the given differential equation: $$y(x) = \arcsin(x) + C_1x + C_2$$

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Integration Techniques
Understanding integration techniques is crucial when solving differential equations. The process of finding the antiderivative, or integral, is often not straightforward and can involve several methods. Sometimes we can use direct integration if the function is a standard form for which we know the antiderivative. However, more complex functions require creative techniques like substitution, integration by parts, partial fraction decomposition, or even numerical methods when an analytical solution is difficult to obtain.

When tackling the integral of a complex function, it's imperative to identify which technique will simplify the problem. For instance, when encountering an integrand with a square root in the denominator, like \( \frac{x}{(1-x^{2})^{\frac{3}{2}}} \), the substitution method often proves useful. In our case, choosing a suitable substitution can help us transform the integrand into a more manageable form and reveal a straightforward path to the solution.
Second Derivative
The second derivative of a function, denoted as \( y''(x) \) or \( \frac{d^{2}y}{dx^2} \), tells us a lot about the function’s curvature. It essentially describes how the rate of change of the function's slope is itself changing. A thorough understanding of second derivatives is vital when solving second-order differential equations.

In the presented exercise, the given equation is a second-order differential equation, indicating that it represents the second derivative of an unknown function \( y(x) \). To solve for \( y(x) \) from its second derivative, we perform integration twice. This process moves us ‘backwards’ from acceleration to velocity (first derivative), and then from velocity to position (the function \( y(x) \)).
Arcsin Function
The arcsin function, also known as the inverse sine function, is a fundamental function in trigonometry. It is the inverse operation of the sine function, designated as \( \arcsin(x) \) or \( \sin^{-1}(x) \). The arcsin function determines the angle whose sine is the given value, which implies it takes an input between -1 and 1, and returns an angle in radians (or degrees) within the interval \( [-\frac{\pi}{2}, \frac{\pi}{2}] \).

In relation to integrals involving square roots and binomials such as \( \frac{1}{\sqrt{1-x^2}} \), knowing that the antiderivative is \( \arcsin(x) \) simplifies the solution process. As showcased in our problem, this knowledge helps us to directly write down part of the general solution for the differential equation.
Substitution Method
The substitution method, often employed in integrating complex functions, involves replacing a part of the integrand with a new variable. This can transform a difficult integral into an easier one. This technique requires two main steps: first, identify an appropriate substitution that simplifies the integral, and second, perform the substitution correctly to find the new limits of integration if necessary.

In our original exercise, we use the substitution \( u = 1 - x^2 \) for the integral of the second derivative. By differentiating this expression with respect to \( x \), we obtain \( du = -2x dx \) which allows us to replace \( x dx \) in the original integrand. This transforms the complicated integral into a simpler form \( u^{-\frac{3}{2}} \), which we can integrate easily. The substitution method is a powerful tool for making seemingly intractable integrals approachable.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Properties of the Gompertz solution Verify that the function 1\(M(t)=K\left(\frac{M_{0}}{K}\right)^{\operatorname{csp}(-n)}\) satisfies the properties \(M(0)=M_{0}\) and \(\lim _{t \rightarrow \infty} M(t)=K\)

Solve the differential equation for Newton's Law of Cooling to find the temperature function in the following cases. Then answer any additional questions. A glass of milk is moved from a refrigerator with a temperature of \(5^{\circ} \mathrm{C}\) to a room with a temperature of \(20^{\circ} \mathrm{C}\). One minute later the milk has warmed to a temperature of \(7^{\circ} \mathrm{C}\). After how many minutes does the milk have a temperature that is \(90 \%\) of the ambient temperature?

Equilibrium solutions \(A\) differential equation of the form \(y^{\prime}(t)=f(y)\) is said to be autonomous (the function \(f\) depends only on y. The constant function \(y=y_{0}\) is an equilibrium solution of the equation provided \(f\left(y_{0}\right)=0\) (because then \(y^{\prime}(t)=0\) and the solution remains constant for all \(t\) ). Note that equilibrium solutions correspond to horizontal lines in the direction field. Note also that for autonomous equations, the direction field is independent of t. Carry out the following analysis on the given equations. a. Find the equilibrium solutions. b. Sketch the direction field, for \(t \geq 0\). c. Sketch the solution curve that corresponds to the initial condition \(y(0)=1\). $$y^{\prime}(t)=\sin y$$

Solving initial value problems Solve the following initial value problems. $$y^{\prime}(t)=t e^{t}, y(0)=-1$$

Solving initial value problems Solve the following initial value problems. $$y^{\prime \prime}(t)=t e^{\prime}, y(0)=0, y^{\prime}(0)=1$$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free