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 November 21st, 2016, 12:50 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???? Thanks from zylo
 November 21st, 2016, 01:13 AM #2 Global Moderator   Joined: Dec 2006 Posts: 19,295 Thanks: 1686 $y = \dfrac{1}{x+C}\implies y' = -\dfrac{1}{(x+C)^2}$, 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? Thanks from hyperbola
 November 21st, 2016, 07:37 AM #3 Senior Member   Joined: Mar 2015 From: New Jersey Posts: 1,400 Thanks: 100 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, 07:50 AM   #4
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Quote:
 Originally Posted by zylo 2) $\displaystyle \frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}=-2y$, harmonic oscillator
Close but no cigar. (And why would you want to smoke one anyway??)

$\displaystyle \frac{dy}{dx} = - y^2 \implies \frac{d^2y}{dx^2} = -2y \frac{dy}{dx}$

-Dan

November 21st, 2016, 08:34 AM   #5
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Quote:
 Originally Posted by topsquark Close but no cigar. (And why would you want to smoke one anyway??) $\displaystyle \frac{dy}{dx} = - y^2 \implies \frac{d^2y}{dx^2} = -2y \frac{dy}{dx}$ -Dan
Yes. But I get half a cigar. A solution of y''+2yy'=0 is also a solution of y'+y$\displaystyle ^{2}$=0, which makes sense since it is non-linear and a solution is not guaranteed unique.

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 08:58 AM.

November 21st, 2016, 08:56 AM   #6
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Quote:
 Originally Posted by zylo A solution of y''+2yy'=0 is also a solution of y'+y$\displaystyle ^{2}$=0, which makes sense since it is non-linear and a solution is not guaranteed unique.
The nonlinearity of the equation has nothing to do with the fact that "A solution of y''+2yy'=0 is also a solution of y'+y$\displaystyle ^{2}$=0".

$y'+y^2=0$ 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 09:05 AM.

November 21st, 2016, 09:16 AM   #7
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Quote:
 Originally Posted by zylo Very interesting thread. Thanks 1) $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}=-y^{2}$ 2) $\displaystyle \frac{\mathrm{d^{2}y} }{\mathrm{d} x^{2}}=-2y\frac{\mathrm{d} y}{\mathrm{d} x}$, corrected by topsquark 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): d(y')=-2ydy y'=-y^{2}+C etc
QUOTE revised

A solution to y'=y$\displaystyle ^{2}$ is not unique since it is non-linear.

 November 21st, 2016, 09:35 AM #8 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,344 Thanks: 2466 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, 09:53 AM   #9
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Quote:
 Originally Posted by zylo A solution to y'=y$\displaystyle ^{2}$ is not unique since it is non-linear.
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, 10:28 AM   #10
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Quote:
 Originally Posted by skipjack 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).
You are kidding.

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