UnkleRhaukus
  • UnkleRhaukus
\[y''-yy'=0\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
the equation is x-absent let \[y'=p\]\[y''=pp'\]
UnkleRhaukus
  • UnkleRhaukus
\[pp'-yp=0\]\[p'-y=0\]\[dp/dx=y\]
UnkleRhaukus
  • UnkleRhaukus
?

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anonymous
  • anonymous
whats your question?
UnkleRhaukus
  • UnkleRhaukus
what do i do next i am tring to solve for y
UnkleRhaukus
  • UnkleRhaukus
y=y(x)
anonymous
  • anonymous
is p a constant?
UnkleRhaukus
  • UnkleRhaukus
p=dy/dx
anonymous
  • anonymous
y'=p =>y''=0
anonymous
  • anonymous
if p is constant
UnkleRhaukus
  • UnkleRhaukus
y''=pp'
anonymous
  • anonymous
p must be a const just check it out!
UnkleRhaukus
  • UnkleRhaukus
that is just the trivial solution
anonymous
  • anonymous
k
UnkleRhaukus
  • UnkleRhaukus
the other solution has tangent of something in it
anonymous
  • anonymous
y'=p y''=p'
anonymous
  • anonymous
ur solving for y right
anonymous
  • anonymous
?
UnkleRhaukus
  • UnkleRhaukus
i am solving for y(x) yes. y'=p however y''=pp'
anonymous
  • anonymous
then you should integrate this equation y'=p => y(x)=px+c
UnkleRhaukus
  • UnkleRhaukus
y(x)=y'x+c
anonymous
  • anonymous
err p is a function of x dhashni
UnkleRhaukus
  • UnkleRhaukus
I am tring to show that the solution is \[y=c\] or \[y=atan({ax \over 2}+b)\]
anonymous
  • anonymous
rather \[y = a \tan(\frac{ax}{2}+b)\]
UnkleRhaukus
  • UnkleRhaukus
um...
UnkleRhaukus
  • UnkleRhaukus
i dont know
anonymous
  • anonymous
aha - it goes tan -> sec^2 -> tan*sec^2 as you differentiate
anonymous
  • anonymous
wait why do you think that y'' = pp' rather y'' = yy'
anonymous
  • anonymous
\[y'' - yy' = 0\]\[y' - \frac{1}{2}y^2 = c\]\[y' = \frac{1}{2}y^2 + c\]\[\frac{y'}{\frac{1}{2}y^2 + c} = 1\]\[\frac{2y'}{y^2 + 2c} = 1\]\[\sqrt{\frac{2}{c}}\arctan(\frac{y}{\sqrt{2c}})=x+c_2\]\[y = \sqrt{2c} \tan(\sqrt{\frac{c}{2}}(x+c_2))\]
anonymous
  • anonymous
the two hard steps are where you integrate both sides with respect to x
UnkleRhaukus
  • UnkleRhaukus
hey thankyou very much
anonymous
  • anonymous
nps :)
UnkleRhaukus
  • UnkleRhaukus
i can not see what you did to \[y''-yy'=0\]to get \[y'-{1 \over 2}y^2=c\]
anonymous
  • anonymous
that's the step where i integrated. the integral of y'' is y'+c and the integral of 2yy' is y^2+c (chain rule)
anonymous
  • anonymous
\[\int (y''-yy') dx = \int 0 \space dx\] \[\int \frac{d}{dx}(y') dx - \frac{1}{2} \int 2y\frac{dy}{dx} dx = 0 + C_3\] \[ y' + C_1 - \frac{1}{2}y^2 - C_2 = C_3 \] \[ y' - \frac{1}{2}y^2 = (-C_1+C_2+C_3) \] \[ y' - \frac{1}{2}y^2 = c \]
UnkleRhaukus
  • UnkleRhaukus
that is it thank you Broken Fixer

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