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UnkleRhaukus
 4 years ago
\[y''yy'=0\]
UnkleRhaukus
 4 years ago
\[y''yy'=0\]

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UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2the equation is xabsent let \[y'=p\]\[y''=pp'\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2\[pp'yp=0\]\[p'y=0\]\[dp/dx=y\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2what do i do next i am tring to solve for y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0p must be a const just check it out!

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2that is just the trivial solution

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2the other solution has tangent of something in it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ur solving for y right

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i am solving for y(x) yes. y'=p however y''=pp'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then you should integrate this equation y'=p => y(x)=px+c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0err p is a function of x dhashni

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2I am tring to show that the solution is \[y=c\] or \[y=atan({ax \over 2}+b)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0rather \[y = a \tan(\frac{ax}{2}+b)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0aha  it goes tan > sec^2 > tan*sec^2 as you differentiate

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait why do you think that y'' = pp' rather y'' = yy'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[y''  yy' = 0\]\[y'  \frac{1}{2}y^2 = c\]\[y' = \frac{1}{2}y^2 + c\]\[\frac{y'}{\frac{1}{2}y^2 + c} = 1\]\[\frac{2y'}{y^2 + 2c} = 1\]\[\sqrt{\frac{2}{c}}\arctan(\frac{y}{\sqrt{2c}})=x+c_2\]\[y = \sqrt{2c} \tan(\sqrt{\frac{c}{2}}(x+c_2))\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the two hard steps are where you integrate both sides with respect to x

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2hey thankyou very much

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2i can not see what you did to \[y''yy'=0\]to get \[y'{1 \over 2}y^2=c\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that'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
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int (y''yy') dx = \int 0 \space dx\] \[\int \frac{d}{dx}(y') dx  \frac{1}{2} \int 2y\frac{dy}{dx} dx = 0 + C_3\] \[ y' + C_1  \frac{1}{2}y^2  C_2 = C_3 \] \[ y'  \frac{1}{2}y^2 = (C_1+C_2+C_3) \] \[ y'  \frac{1}{2}y^2 = c \]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.2that is it thank you Broken Fixer
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