/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 If \(y=e^{\tan ^{-1}} x\), show ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 8

If \(y=e^{\tan ^{-1}} x\), show that $$ \left(x^{2}+1\right) \frac{d^{2} y}{d x^{2}}+(2 x-1) \frac{d y}{d x}=0 $$

Short Answer

Expert verified
After computing the 1st and 2nd derivative of given \( y \), when they're substituted into the given equation, we get \(0 = 0\), which is always true. Hence, the given differential equation holds for \( y = e^{\tan ^{-1} x} \).

Step by step solution

01

Compute the first derivative of \( y \)

Start by finding the first derivative of \( y \) with respect to \( x \) using the chain rule, we get \[ \frac{d y}{d x} = e^{\tan ^{-1} x} \cdot \frac{1}{1+x^2} = \frac{y}{1+x^2}\]
02

Compute the second derivative of \( y \)

Now, compute the second derivative of \( y \) with respect to \( x \) by differentiating the result from step 1 with respect to \( x \):\[ \frac{d^{2} y}{d x^{2}} = \frac{(1+x^2)\cdot y' - y\cdot 2x}{(1+x^2)^2} \] Substitute \( y' \) from step 1 in to get \[ \frac{d^{2} y}{d x^{2}} = \frac{(1+x^2)\cdot y/(1+x^2) - y\cdot 2x}{(1+x^2)^2} = \frac{y - 2xy}{(1+x^2)^2} \]
03

Verify if the equation holds

Next, you substitute \( y \), \( y' \), and \( y'' \) into the given equation to check if it holds. \[ (x^{2}+1) \cdot \left(y - 2xy\right)/(1+x^2)^2 +(2x-1) \cdot y/(1+x^2) \] Simplifying, we receive:\[ y/(1+x^2) - 2xy/(1+x^2) + 2xy/(1+x^2) - y/(1+x^2)=0 \] Observe that the equation becomes \( 0 =0 \), which holds true for every \(x\). Thus, the given differential equation has been verified.

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 91影视!

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

If \(\tan ^{-1}\left(\frac{x^{2}-y^{2}}{x^{2}+y^{2}}\right)=a\) prove that \(\frac{d y}{d x}=\frac{x(1-\tan a)}{y(1+\tan a)}\).

\begin{aligned} &\text { If } f(x)=x+\tan x \text { and } g \text { is the inverse of } f \text {, then }\\\ &\text { prove that } g^{\prime}(x)=\frac{1}{2+\tan ^{2}(g(x))} . \end{aligned}

If \(y=\left(C_{1}+C_{2} x\right) \sin x+\left(C_{3}+C_{4} x\right) \cos x\), show that \(\frac{d^{4} y}{d x^{4}}+2 \frac{d^{2} y}{d x^{2}}+y=0\)

If \(x=a \cos \theta, y=b \sin \theta\), find \(\frac{d^{2} y}{d x^{2}}\)

If \(f(x)=a x+b, x \in[-2,2]\), then the point \(c \in(-2,2)\) where $$ f(c)=\frac{f(2)-f(-2)}{4} $$ (a) does not exist (b) can be any \(c \in(-2,2)\) (c) can be only 1 (d) can be only \(-1\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Probability and Statistics

Read Explanation

Statistics

Read Explanation
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.