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November 21st, 2016, 12:50 AM   #1
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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????
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November 21st, 2016, 01:13 AM   #2
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$y = \dfrac{1}{x+C}\implies y' = -\dfrac{1}{(x+C)^2}$, 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?
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November 21st, 2016, 07:37 AM   #3
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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) ?
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November 21st, 2016, 07:50 AM   #4
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Originally Posted by zylo View Post
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
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November 21st, 2016, 08:34 AM   #5
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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
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Last edited by zylo; November 21st, 2016 at 08:58 AM.
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November 21st, 2016, 08:56 AM   #6
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Originally Posted by zylo View Post
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.
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November 21st, 2016, 09:16 AM   #7
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Quote:
Originally Posted by zylo View Post
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.
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November 21st, 2016, 09:35 AM   #8
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You do realise that 1, 2 and 3 are all just derivatives and integrals of each other? They are essentially the same equation.
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November 21st, 2016, 09:53 AM   #9
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Quote:
Originally Posted by zylo View Post
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).
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November 21st, 2016, 10:28 AM   #10
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Quote:
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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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