Solution for 2nd ODE Question 2d2ydx2 + 4 dydx+ 7y = e^−x cos x

[song90]

2d2ydx2 + 4 dydx+ 7y = e^−x cos x

i solved the equation for yc. but couldn't solve for yp as i dint know what kind of undetermined method to put for yp, particular integral.

 

[Mark44]

2d2ydx2 + 4 dydx+ 7y = e^−x cos x

i solved the equation for yc. but couldn't solve for yp as i dint know what kind of undetermined method to put for yp, particular integral.

Your equation would be more readable as
2y'' + 4y' + 7y = e^-xcos x

For your particular solution, try y_p = e^-x(Acos x + Bsin x).

 

[snipez90]

Try e^-x (A*cos x + B*sin x)

 

[song90]

Your equation would be more readable as
2y'' + 4y' + 7y = e^-xcos x

For your particular solution, try y_p = e^-x(Acos x + Bsin x).

do e^{-x there put a constant?
let say _p=Ce-x(Acos x + Bsin x)}

 

[tiny-tim]

hi song90!

you can absorb the C into the other two constants

ie Ce^-x(Acos x + Bsin x) = e^-x(ACcos x + BCsin x),

so just rename AC A and BC B

 

[song90]

sorry, i am new to this forum...
yp = Ce-x(Acos x + Bsin x). insert a constant at e-x since it is one of the members of undetermined coefficient?

2y'' + 4y' + 7y = e-xcos x <this is the original question and the answer for yp is 1/3(e-xcos x). Still i could not manage to get the ans for yp. Looking for help..

 

[song90]

thanks for immediate reply. i was trying to solve the equations but i still couldn't get the answer for yp which is 1/3(e-xcos x). Wondering where the mistakes i had made

 

[tiny-tim]

sorry, i am new to this forum...
yp = Ce-x(Acos x + Bsin x). insert a constant at e-x since it is one of the members of undetermined coefficient?

i'm not following you

(and anyway, it makes no difference, as i said before)

2y'' + 4y' + 7y = e-xcos x <this is the original question and the answer for yp is 1/3(e-xcos x). Still i could not manage to get the ans for yp. Looking for help..

that's A = 1/3, B = 0, why couldn't you get it?

show us what you've tried ​

 

[song90]

okay looked back on my workings i found that i had timed a "x" in the yp equation,
yp=(e-x(ACcos x + BCsin x))x. i was thinking yp had common with yc as
yc=e-x(Asin√(5/2)+Bcos√(5/2))

 

[tiny-tim]

sorry, I've no idea what you're doing

put y = e^-x(Acos x + Bsin x)

into 2y'' + 4y' + 7y = e^-xcos x

to find A and B

 

[song90]

i was told by lecturer (maybe i misunderstanding what she said). If yp has the common with yc, yp=(equation)x. something like this... I have no idea about this lor.

anyways thanks a lot for your help.

 

[tiny-tim]

i was told by lecturer (maybe i misunderstanding what she said). If yp has the common with yc, yp=(equation)x. something like this...

i think she meant that if e^-x or cosx was a solution for y_c (in this case, it isn't),

then you would need to use xe^-x(Acosx + Bsinx) instead of just e^-x(Acosx + Bsinx)

 

[Mark44]

i think she meant that if e^-x or cosx was a solution for y_c (in this case, it isn't),

then you would need to use xe^-x(Acosx + Bsinx) instead of just e^-x(Acosx + Bsinx)

Yes, I think the OP was trying to say this.