math help!!! (diff eq.)

[Suspicious-Teach8788]

y'' +2y = x
y(0) = 0, y(pi) = 0

Thanks!

 

[chuckywang]

Originally posted by: DLeRium
y'' +2y = x
y(0) = 0, y(pi) = 0

Thanks!

Hold on...

 

[Suspicious-Teach8788]

I was here last week asking for help when its like y'' + 4y = cos x.. I understand those now.. like how to guess that y = A cos x or whatever.. but when the right side = something funky, I'm clueless =(

 

[chuckywang]

Are you sure it's y''+2y?

 

[Suspicious-Teach8788]

yup... ok so I got it I think.. you guess y = x/2

 

[chuckywang]

Originally posted by: DLeRium
yup... ok so I got it I think.. you guess y = x/2

Yes, but that's just the particular part. You still need the homogenous part of the solution (with the two constants you solve for with the initial conditions). One of the constants turns out to be equal to zero, but the other one is something really nasty.

 

[Suspicious-Teach8788]

yup thats correct.. it is nasty.. the answer freaked me out at first, but its not all THAT bad... -c2 = 2 / sin(root 2 * pi)

 

[chuckywang]

Originally posted by: DLeRium
yup thats correct.. it is nasty.. the answer freaked me out at first, but its not all THAT bad... -c2 = pi / (2* sin(root 2 * pi))

That's what I got.

 

[cirthix]

Originally posted by: DLeRium
y'' +2y = x
y(0) = 0, y(pi) = 0

Thanks!

here's what i get:

try y=ax+b
substitung into original equation: 0+2ax+2b=x
a=1, b=0
PSDE: y=x/2


HDE: y''+2y=0
CP: r²+2
roots: 0+/- isqurt(2)
GSHDE: y=Ccos(sqrt2)x+Dsin(sqrt2)x
using the inital values, y(0)=Ccos(0)+Dsin(0)=0
and so, C+0D=0, c=0
GSHDE: y=Dsin(sqrt2)x
using the initial values, y(pi)=Dsin(sqrt2)pi=0
and so, D=0

Solution to the IVP: y=x/2

dam, i just noticed this doesnt hold true for the second initial condition, not sure where i wetnt wrong. hmm...

 

[chuckywang]

Originally posted by: cirthix

Originally posted by: DLeRium
y'' +2y = x
y(0) = 0, y(pi) = 0

Thanks!

here's what i get:

try y=ax+b
substitung into original equation: 0+2ax+2b=x
a=1, b=0
PSDE: y=x/2


HDE: y''+2y=0
CP: r²+2
roots: 0+/- isqurt(2)
GSHDE: y=Ccos(sqrt2)x+Dsin(sqrt2)x
using the inital values, y(0)=Ccos(0)+Dsin(0)=0
and so, C+0D=0, c=0
GSHDE: y=Dsin(sqrt2)x
using the initial values, y(pi)=Dsin(sqrt2)pi=0
and so, D=0

Solution to the IVP: y=x/2

dam, i just noticed this doesnt hold true for the second initial condition, not sure where i wetnt wrong. hmm...

You went wrong cause you didn't include the particular solution when solving for the constants.

 

[cirthix]

Originally posted by: chuckywang

Originally posted by: cirthix

Originally posted by: DLeRium
y'' +2y = x
y(0) = 0, y(pi) = 0

Thanks!

here's what i get:

try y=ax+b
substitung into original equation: 0+2ax+2b=x
a=1, b=0
PSDE: y=x/2


HDE: y''+2y=0
CP: r²+2
roots: 0+/- isqurt(2)
GSHDE: y=Ccos(sqrt2)x+Dsin(sqrt2)x
using the inital values, y(0)=Ccos(0)+Dsin(0)=0
and so, C+0D=0, c=0
GSHDE: y=Dsin(sqrt2)x
using the initial values, y(pi)=Dsin(sqrt2)pi=0
and so, D=0

Solution to the IVP: y=x/2

dam, i just noticed this doesnt hold true for the second initial condition, not sure where i wetnt wrong. hmm...

You went wrong cause you didn't include the particular solution when solving for the constants.

doh! you're right. i do that pretty often