# Differential equation problem

1. Jun 7, 2005

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?

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

2. Jun 7, 2005

### Hurkyl

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

3. Jun 7, 2005

### 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.

4. Jun 7, 2005

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

5. Jun 7, 2005

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?

6. Jun 7, 2005

### dextercioby

You have this system

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

Solve it.

Daniel.

7. Jun 7, 2005

i don't see why could you elaborate?

8. Jun 7, 2005

### dextercioby

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

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

Daniel.

9. Jun 7, 2005