Need help with solving Non-homogenous 2Nd ODE Eqn.

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Dec 16th 2010, 10:50 PM
saykhong

Need 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 PM
kjchauhan

$y'' + 7y' -8y = e^x$

Auxiliary equation is

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

$\therefore D=-8, 1$

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

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

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

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

Now,

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

then substitute x= o in last both result and solve for $c_1, c_2$
Dec 16th 2010, 11:45 PM
FernandoRevilla

These kind of problems are just routine knowing the corresponding theory. What difficulties have you had?.

Fernando Revilla
Dec 16th 2010, 11:49 PM
saykhong

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 AM
HallsofIvy

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

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

Of course, you are still not finished. Now find $c_1$ and $c_2$ so that y(0)= 8/9 and y'(0)= 1.
Dec 17th 2010, 05:08 AM
saykhong

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 AM
Prove It

No, since $\displaystyle e^{-2x}$ is not a term that appears in your homogeneous solution. You can just choose $\displaystyle Ae^{-2x}$ as the nonhomogeneous solution.