My Math Forum A beginner differential equation question

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 June 17th, 2017, 12:32 PM #1 Member     Joined: Jun 2017 From: India Posts: 65 Thanks: 3 A beginner differential equation question What is a differential equation ? It is an equation involving an unknown function (solution) and its derivatives .
 June 17th, 2017, 12:51 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,966 Thanks: 2215 The function (of x) is y, and dy/dx is its derivative (with respect to x). Thanks from awholenumber
 June 17th, 2017, 04:33 PM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 If you are asking whether $\displaystyle y= \pm\sqrt{\frac{2}{3}x^3+ C}$ really does satisfy $\displaystyle \frac{dy}{dx}= \frac{x^2}{y}$, it is easy to check. First, squaring both sides gives $\displaystyle y^2= \frac{2}{3}x^3+ C$ and differentiating both sides with respect to x, $\displaystyle 2y\frac{dy}{dx}= 2x^2$. Dividing both sides by 2y, $\displaystyle \frac{dy}{dx}= \frac{x^2}{y}$, precisely the differential equation you started with. Or, more directly but more complicated, differentiating the original formula with respect to x, $\displaystyle \frac{dy}{dx}= \pm\frac{1}{2}\left(\frac{2}{3}x^3+ C\right)^{-1/2}(2x^2)= \frac{x^2}{\pm\sqrt{\frac{2}{3}x^3+ C}}= \frac{x^2}{y}$. Thanks from awholenumber Last edited by skipjack; June 17th, 2017 at 05:16 PM.
 June 17th, 2017, 10:17 PM #4 Member     Joined: Jun 2017 From: India Posts: 65 Thanks: 3 Thanks , Found some good materials here National Council Of Educational Research And Training :: Home

 Tags beginner, differential, equation, question

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post colinzeal Algebra 5 September 30th, 2013 03:35 PM hogan Differential Equations 4 June 4th, 2012 11:02 AM phss Calculus 2 January 19th, 2011 10:08 AM and1bball4mk Differential Equations 2 January 13th, 2010 07:10 AM Kevlarji Differential Equations 1 September 26th, 2009 03:25 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top