Dif. eq. problem - dont know why it is wrong

[UrbanXrisis]

dif. eq. problem -- don't know why it is wrong

I did this problem three times with the same wrong answer...

solve for y:
$$y'''-9y''+18y'=0$$
y(0)=5
y'(0)=7
y''(0)=6

so I change it into the auxiliary equation then solve for the variables, i get the roots: 3 and 6. so the general solution would be:

$$y'=c_1e^{3x}+c_2e^{6x}$$
solving for c1 and c2 i get:

$$7=c_1+c_2$$
$$2=c_1+2c_2$$
$$c_1=12$$
$$c_2=-5$$

$$y=\int{12e^{3x}-5e^{6x}}$$
$$y=36e^{3x}-30e^{6x}+c$$

solving for c when x=0 and y=5, i get

$$y=36e^{3x}-30e^{6x}-1$$

what am i donig wrong?

 

[HallsofIvy]

Hey! y'''- 9y"+ 18y'= 0 is a third order differential equation!
Its auxiliary equation, r³- 9r²+ 18r= r(r²- 9r+ 18)= r(r- 3)(r- 6) has three roots! They are 0, 3, and 6.

The general solution for y is y(x)= C₁e^3x+ C₂e^6x+ C₃ (since e^0x= e⁰= 1). Use that to satisfy the initial conditions:
y(0)= C₁+ C₂+ C₃= 5. y'(0)= 3C₁+ 6C₂= 7, and y"(0)= 9C₁+ 36C₂= 6.

That's how I would have done it. What you did was treat it as a second order d.e. for y' and then integrate. You would have gotten exactly right answer except the anti-derivative of e^a is (1/a)e^ax, not ae^ax!

 

[jamesrc]

Double check your integration:

$$\int 12 e^{3x}=12\frac{e^{3x}}{3}+c$$

and so on...