y''+7y'-8y = e^x y(0)= 8/9 y'(0)=1

Working will be much appreciated.

Thanks a million!

Say Khong.

God Bless.

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- Dec 16th 2010, 11:50 PMsaykhongNeed help with solving Non-homogenous 2Nd ODE Eqn.
y''+7y'-8y = e^x y(0)= 8/9 y'(0)=1

Working will be much appreciated.

Thanks a million!

Say Khong.

God Bless. - Dec 17th 2010, 12:43 AMkjchauhan

Auxiliary equation is

Now P.I. =

Now,

then substitute x= o in last both result and solve for - Dec 17th 2010, 12:45 AMFernandoRevilla
These kind of problems are just routine knowing the corresponding theory. What difficulties have you had?.

Fernando Revilla - Dec 17th 2010, 12:49 AMsaykhong
Apparently the answer doesn't end after finding the auxiliary eqn, the eqn still need to be solve for y''+7y'-8y = e^x, which is the particular solution of that eqn.

- Dec 17th 2010, 01:46 AMHallsofIvy
Yes, and Kjchauhan did that rather "offhandedly" with " ". Quite a jump!

You can solve for the particular solution by "variation of parameters" or, more simply in this case, "undetermined coefficients". The right hand side of the equation is so normally, we would try something of the form . But we have already seen that was a solution to the associated homogeneous equation so there will be no value of A that will give anything but "0" on the left.

Instead try . Then and . Putting those into the equation gives so we can take .

That gives as kjchauhan said.

Of course, you are**still**not finished. Now find and so that y(0)= 8/9 and y'(0)= 1. - Dec 17th 2010, 06:08 AMsaykhong
Erm,Say for example, if the equation ends up with y''+7y'-8y = e^(-2x), Does it mean that I will assume form y to be Axe^(-2x)

- Dec 17th 2010, 08:09 AMProve It
No, since is not a term that appears in your homogeneous solution. You can just choose as the nonhomogeneous solution.