Second order differential equation using double substitutionWatch

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#1
x(d^2y/dx^2) + x(dy/dx)^2 + 0.5(dy/dx)=0.25 hint use x=t^2 , u=e^y then derive a differential equation for u(t)
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3 years ago
#2
(Original post by acidkc111)
x(d^2y/dx^2) + x(dy/dx)^2 + 0.5(dy/dx)=0.25 hint use x=t^2 , u=e^y then derive a differential equation for u(t)
are you posing this as a challenge?
0
#3
I completed all my work and I asked my professor for a bonus question. Just a hint on how you'd go about solving with the substitution would help
0
3 years ago
#4
(Original post by acidkc111)
I completed all my work and I asked my professor for a bonus question. Just a hint on how you'd go about solving with the substitution would help
the hint given is sufficient

use the first substitution first then the second
(However this is not a sensible way of solving this ... I agree with the first substitution then I would use something which very obvious)
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#5
(Original post by TeeEm)
the hint given is sufficient

use the first substitution first then the second
(However this is not a sensible way of solving this ... I agree with the first substitution then I would use something which very obvious)
would you differentiate x=2t with respect to u so (dx/du)=2(dt/du) then have (dy/dx)=(dy/du)*(du/dx) where (dy/du)=e^-y so (dy/dx)=(1/2)*(e^-y)(du/dt)
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3 years ago
#6
Have you covered substitutions when it comes to ODEs?
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#7
yes but im not sure what to do when I am given 2 hints it confuses me more than anything
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#8
i used the substitution x=t^2 and got (dx/dt)=2t (dy/dx)=(dy/dt)(dt/dx) (dy/dx)=(dy/dt)(1/2t) (d^2y/dx^2)=((d^2y/dt^2)*1/2t -1/2t^2(dy/dt))1/2t subbing this in and cancelling I got

d^2y/dt^2 +(dy/dt)^2=1
1
3 years ago
#9
(Original post by acidkc111)
i used the substitution x=t^2 and got (dx/dt)=2t (dy/dx)=(dy/dt)(dt/dx) (dy/dx)=(dy/dt)(1/2t) (d^2y/dx^2)=((d^2y/dt^2)*1/2t -1/2t^2(dy/dt))1/2t subbing this in and cancelling I got

d^2y/dt^2 +(dy/dt)^2=1
correct

now use the other substitution (easier)
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#10
(Original post by TeeEm)
correct

now use the other substitution (easier)
Should I use the substitution from where I am at now or from the beginning and what should I differentiate u=e^y with respect to
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3 years ago
#11
(Original post by acidkc111)
Should I use the substitution from where I am at now or from the beginning and what should I differentiate u=e^y with respect to
from where you are now
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#12
I wrote u=e^y as lnu=y -> 1/u = dy/du dy/dt = (u^-1 du/dt) d^2y/dt^2 =(-u^-2)(du/dt)^2 +(u^-1)(d^2u/dt^2) subbing in

d^2u/dt^2 -u=0 the Aux equation is m^2-1=0 m=1 m=-1 so u(t)=Ae^t+Be^-t
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