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.

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

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 $\displaystyle c_1, c_2$

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

Fernando Revilla

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.

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.

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)

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.