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, 10:50 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 16th 2010, 11:43 PMkjchauhan
$\displaystyle y'' + 7y' -8y = e^x$

Auxiliary equation is

$\displaystyle D^2+7D-8=0$

$\displaystyle \therefore D=-8, 1$

$\displaystyle \therefore C.F.= c_1 e^{-8x} + c_2 e^{x}$

Now P.I. = $\displaystyle \frac{1}{D^2+7D-8} e^x$ $\displaystyle =\frac{1}{9}xe^x$

$\displaystyle \therefore y(x) = C.F. + P.I.$

$\displaystyle \therefore y(x) = c_1 e^{-8x} + c_2 e^{x} + \frac{1}{9} xe^x$

Now,

$\displaystyle y'(x) = -8 c_1 e^{-8x} + c_2 e^{x} + \frac{1}{9} xe^x + \frac{1}{9} e^x$

then substitute x= o in last both result and solve for $\displaystyle c_1, c_2$ - Dec 16th 2010, 11:45 PMFernandoRevilla
These kind of problems are just routine knowing the corresponding theory. What difficulties have you had?.

Fernando Revilla - Dec 16th 2010, 11:49 PMsaykhong
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, 12:46 AMHallsofIvy
Yes, and Kjchauhan did that rather "offhandedly" with "$\displaystyle \frac{1}{D^2+7D-8} e^x= \frac{1}{9}xe^x$". 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 $\displaystyle e^x$ so normally, we would try something of the form $\displaystyle y= Ae^x$. But we have already seen that $\displaystyle e^x$ 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 $\displaystyle y= Axe^x$. Then $\displaystyle y'= Ae^x+ Axe^x$ and $\displaystyle y"= 2Ae^x+ Axe^x$. Putting those into the equation gives $\displaystyle y"+ 7y'- 8y= 2Ae^x+ Axe^x+ 7Ae^x+ 7Axe^x- 8Ae^x= (2Ae^x+ 7Ae^x)+ (Axe^x+ 7Axe^x- 8Axe^x)= 9Ae^x$ so we can take $\displaystyle A= \frac{1}{9}$.

That gives $\displaystyle y(x)= c_1e^x+ c_2e^{-8x}+ \frac{1}{9}xe^x$ as kjchauhan said.

Of course, you are**still**not finished. Now find $\displaystyle c_1$ and $\displaystyle c_2$ so that y(0)= 8/9 and y'(0)= 1. - Dec 17th 2010, 05:08 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, 07:09 AMProve It
No, since $\displaystyle \displaystyle e^{-2x}$ is not a term that appears in your homogeneous solution. You can just choose $\displaystyle \displaystyle Ae^{-2x}$ as the nonhomogeneous solution.