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For the differential equation $\frac{dy}{dx}=\sqrt{y^2-25}$ does the existence/uniqueness theorem guarantee that there is a solution to this equation through the point

  1. (-1,28)

  2. (0,5)

  3. (3,-5)

  4. (1,34)

given that $\frac{dy}{dx}=f(x,y)=\sqrt{y^2-25}$

$f'=\frac{2y}{\sqrt{y^2-25}}$

this means as existence and uniqueness theorem the interval containing 5 is not have solution i am right

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On solving the differential equation, we get $$\ln|y+\sqrt{y^2-25}|=x+c$$ Now if you put the $4$ points here, do you get real finite values of $c$?

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  • $\begingroup$ ,...yes all points give finite value of c $\endgroup$ – user271336 Feb 19 '17 at 11:32
  • $\begingroup$ @Psuresh So your solution exists for each point. $\endgroup$ – SchrodingersCat Feb 19 '17 at 11:33
  • $\begingroup$ ....3 point getting negative sing of ln is that mean ther is no solution exist $\endgroup$ – user271336 Feb 19 '17 at 11:41
  • $\begingroup$ Don't forget the modulus sign .. you only take the absolute value. $\endgroup$ – SchrodingersCat Feb 19 '17 at 11:46
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You can easily compute the complete solution by first noticing that $y\equiv \pm 5$ are constant solutions and that under the substitution $y=5\cosh(u(x))$ for $y>0$ the equation transforms to $$ y'=5\sinh(u(x))u'(x)=5|\sinh(u(x))| $$ so that $u(x)=x+c$ for $u(x)\ge 0$. This allows to patch solutions together, $$ y(x)=\begin{cases} 5&x<c,\\ 5\cosh(x-c)&x\ge c. \end{cases} $$ Similarly for $y<0$ one finds the solutions $$ y(x)=\begin{cases} -5\cosh(x-c)&x< c,\\ -5&x\ge c. \end{cases} $$

In short, if the task also includes the "uniqueness" part of the existence and uniqueness theorem, then you get a different selection of choices that apply.

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