Solution to (D+2)(D+3)y=4t+5e^t with Initial Conditions y(0)=4, y'(0)=5

[RadiationX]

Solve the following differential equation:

(D + 2)(D + 3)y = 4t + 5e^t; y(0)=4, y'(0)=5

I have the following as the answer is it correct?

y=17e^{-2t} -13e^{-3t} + \frac{4}{11}t + \frac{5}{12}e^t

 

[Hurkyl]

You try plugging your answer back into the equation, to see if it works out right?

 

[dextercioby]

Nope.Here's what Maple gives as the general solution

\frac{d^2 y}{dx^2}+5\frac{dy}{dx}+6y=4x+5e^x Exact solution is :

y\left( x\right) =\frac{2}{3}x-\frac{5}{9}+\frac{5}{12}e^x+C_1 e^{-3x}+C_2 e^{-2x}

I get some nasty looking coeff.

Daniel.

 

[RadiationX]

i am way off then. by the way could MATLAB solve this problem?

 

[RadiationX]

I'm not understanding what my professor calls an annihilator. what i have is

this D.E.

(D + 2)(D + 3)y = 4t + 5e^t;y(0)=4,y'(0)=5

I need to find a differential opperator that will make the right side of this D.E.

zero. So i think that this differential opperator (annihilator) should be this.

D^2(D-1)

So to solve this problem i need to solve this homogenous D.E. first:

(D + 2)(D + 3)y=0 and its solutions are this:

y_c=C_1e^{-2t} + C_2e^{-3t}

now i need to find the other part which involves the differential opperator. so

now i need to solve this part D^2(D-1)=0

which yields C_3 + tC_4 + C_5e^t

if all of the above is correct i need to find out what the arbitrary constants of

C_3 + tC_4 + C_5e^t are

so what i do is say that this C_3 + tC_4 + C_5e^t is equal to

y_p

so now i have y_p= C_3 + tC_4 + C_5e^t

now i find y'_p and y''_p

y'_p= C_4 +C_5e^t

y''_p=C_5e^t

now what i do is plug in all the y-sub-p's into this y'' + 5y' +6y= 4t +5e^t

and match up the the coef. to find out what the constants are

what i get is this

12C_5e^t + 5C_4 + 6tC_4 + 6C_3= 4t + 5e^t

now my problem is that the C-sub-4's are not the same. one of them is

multiplied by t which is a problem because i can't get the coef. to match

what have i done wrong?

 

[dextercioby]

You have this system

\left\{\begin{array}{c} 5C_{4}+6C_{3}=0\\6C_{4}=4 \end{array} \right

Solve it.

Daniel.

 

[RadiationX]

You have this system

\left\{\begin{array}{c} 5C_{4}+6C_{3}=0\\6C_{4}=4 \end{array} \right

Solve it.

Daniel.

i don't see why could you elaborate?

 

[dextercioby]

12C_5e^t + 5C_4 + 6tC_4 + 6C_3= 4t + 5e^t

That should read

12C_{5} e^{t}+6C_{4}t+\left(5C_{4}+6C_{3}\right)\equiv 5e^{t}+4t+0

Do you see where the system i posted comes from...?

Daniel.

 

[RadiationX]

ahh! yes i see now. i made a simple mistake, but to be truthful i did not know that i could combine csub4 and csub3 but now i see why i can.