find y as a function of x

[moy1989]

Okay guys I am having much difficulty with this problem:

find y as a function of x if y''' - 12y'' + 35y' = 0, y(0) = 1, y'(0) = 2, y''(0) = 5

I started out the problem by writing the equation in characteristic form:

r^3 - 12r^2 + 35r = 0;

I know one of the roots is 0, but how do I find the other two roots. I've searched for factoring methods, but I don't know how to factor it.
Also, on my graphing calculator the curve only shows one root, which is 0. Is this the only root, or are there imaginary roots I'm supposed to look for, if that makes sense?

Please help.

 

[galactus]

With all due respect, you're in a DE class and you can't factor?. That ought to be second nature by now.

Factor the auxiliary equation you have:

\(\displaystyle r(r-7)(r-5)\)

As we can see, the values are 0, 5, 7.

\(\displaystyle C_{1}+C_{2}e^{5t}+C_{3}e^{7t}\)

You can use the initial conditions to find the C values.

 

[Deleted member 4993]

Okay guys I am having much difficulty with this problem:

find y as a function of x if y''' - 12y'' + 35y' = 0, y(0) = 1, y'(0) = 2, y''(0) = 5

I started out the problem by writing the equation in characteristic form:

r^3 - 12r^2 + 35r = 0;

Just looking at a cubic scared you out of your wits!!!

r[sup:3ohhwxr9]3[/sup:3ohhwxr9] - 12r[sup:3ohhwxr9]2[/sup:3ohhwxr9] + 35r = 0

r * (r[sup:3ohhwxr9]2[/sup:3ohhwxr9] - 12r + 35r) = 0

The term within the parenthesis is a quadratic - surely you know how to factorize that. Even if it comes out as complex root remeber Euler's equation

e[sup:3ohhwxr9]ix[/sup:3ohhwxr9] = cos(x) + i sin(x)

and you are on....


Please help.