November 21st, 2016, 01:50 AM  #1 
Senior Member Joined: Dec 2014 From: The Asymptote Posts: 142 Thanks: 6 Math Focus: Certainty  Differential Equations
I require some clarification please, Given $\displaystyle y' = y^2$ Verify that all members of $\displaystyle y = \frac{1}{x+C}$* are solutions of $\displaystyle y'$ $\displaystyle y' = \frac{dy}{dx} = \frac{d}{dx}\bigg(\frac{1}{x+C}\bigg) = \frac{1}{(x+C)^2}$ 1) $\displaystyle y = \frac{1}{x+C}$ 2) Substituting $\displaystyle y$ into $\displaystyle y'$ $\displaystyle y' = y^2 = \bigg(\frac{1}{x+C}\bigg)^2 = \frac{1}{(x+C)^2}$ $\displaystyle y' = y^2$ holds true therefore $\displaystyle y$ is a solution of $\displaystyle y'$ I'm then asked to find a solution of $\displaystyle y' = y^2$ that is not a family member of $\displaystyle y$* ???? Lastly, a solution of the initial value problem: $\displaystyle y' = y^2$ $\displaystyle y(0) = 0.5$ The solution is $\displaystyle y = \frac{1}{x+2}$… in application this could be the vertical displacement of an oscillating spring at time t = 0???? 
November 21st, 2016, 02:13 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,100 Thanks: 1905 
$y = \dfrac{1}{x+C}\implies y' = \dfrac{1}{(x+C)^2}$, so $y' = y^2$ is satisfied. Another solution of $y' = y^2$ is $y = 0$, which is where you're from. On reaching the end of the question, why did you mention an oscillating spring, which seems to have no relevance or connection? 
November 21st, 2016, 08:37 AM  #3 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119 
Very interesting thread. Thanks 1) $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=y^{2}$ 2) $\displaystyle \frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}=2y$, harmonic oscillator 3) $\displaystyle \frac{\mathrm{d} (\frac{\mathrm{d} y}{\mathrm{d} x}y+y^{2})}{\mathrm{d} x}=0$ By 3), a solution of 2) satisfies 1). Solution of 2): 4) y=Acoswx+Bsinwx, w$\displaystyle ^{2}$=2 But I can't make 4) a solution of 1) ? 
November 21st, 2016, 08:50 AM  #4 
Math Team Joined: May 2013 From: The Astral plane Posts: 1,980 Thanks: 789 Math Focus: Wibbly wobbly timeywimey stuff.  
November 21st, 2016, 09:34 AM  #5  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119  Quote:
The thanks was a painful one, since I visited to correct myself. EDIT: Actually, y''+2yy'=0 is solvable: d(y')=2ydy y'=y^{2}+C etc REF: y"+yy'=0 Last edited by zylo; November 21st, 2016 at 09:58 AM.  
November 21st, 2016, 09:56 AM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,559 Thanks: 2558 Math Focus: Mainly analysis and algebra  Quote:
$y'+y^2=0$ is separable and thus completely solvable. The $y=0$ solution appears separately because we divide by $y^2$ in separating the variables, thus denying $y=0$. We thus have to treat that case separately. The direction field will confirm that there are no other solutions. Last edited by v8archie; November 21st, 2016 at 10:05 AM.  
November 21st, 2016, 10:16 AM  #7  
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119  Quote:
A solution to y'=y$\displaystyle ^{2}$ is not unique since it is nonlinear.  
November 21st, 2016, 10:35 AM  #8 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,559 Thanks: 2558 Math Focus: Mainly analysis and algebra 
You do realise that 1, 2 and 3 are all just derivatives and integrals of each other? They are essentially the same equation.

November 21st, 2016, 10:53 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 20,100 Thanks: 1905  Nonsense. If one is given that $y' = y^2$ for all values of $x$, $y = \dfrac{1}{x+C}$ isn't a solution, as it doesn't define $y$ when $x = C$, but $y = 0$ is a solution (the only, and therefore unique, solution).

November 21st, 2016, 11:28 AM  #10 
Senior Member Joined: Mar 2015 From: New Jersey Posts: 1,637 Thanks: 119  

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