Initial value problem

1. Feb 27, 2005

p53ud0 dr34m5

$$\frac{dy}{dx}=x-4-xy-4y$$
you are given that $y(0)=4$.

so, here's what i did:
$$\frac{dy}{dx}=(x-4)(y+1)$$

$$\frac{dy}{y+1}=(x-4)dx$$

i integrated both sides:
$$ln(y+1)=\frac{x^2}{2}-4x+C$$

$$y=e^{\frac{x^2}{2}}e^{-4x}e^{C}-1$$

plugged in for x and y:
$$4=e^{\frac{0^2}{x}}e^{-4*0}e^{C}-1$$

$$5=e^C$$

so:
$$y=5e^{\frac{x^2}{x}}e^{-4x}$$

i know that's wrong, and i need help working it out. that's what i have so far though.

2. Feb 27, 2005

dextercioby

The initial formula (when u restrained into a product of quantities in round brackets) is incorrect.

You may wanna check again.

Daniel.

3. Feb 27, 2005

HallsofIvy

Staff Emeritus
In other words, you factored wrong

x- 4- xy- 4y is not (x- 4)(y+1)= x- 4+ xy- 4y.

4. Feb 27, 2005

dextercioby

The bad news that the solution to this Cauchy problem,according to my version of Maple,is not pretty,not pretty at all...

Are u sure this is the equation u were supposed to solve...?

Daniel.

5. Feb 27, 2005

p53ud0 dr34m5

nah, i didnt factor wrong. i just wrote down the equation wrong.
$$\frac{dy}{dx}=x-4+xy-4y$$

try that one?

6. Feb 27, 2005

dextercioby

It's something else.In this case,your initial factoring turns out to be correct (in this case!) and so the solution is obtained the way you did.Unfortunately,after computing that integration constant,when writing the final solution,you made a mistake.Can u "fix" it...?

Daniel.

7. Feb 27, 2005

p53ud0 dr34m5

$$y=5e^{\frac{x^2}{x}}e^{-4x}-1$$
maybe? haha, im fairly tired and at work. im missing a lot of things.

8. Feb 27, 2005

dextercioby

It's a typo (again!),it's not "x",but "2" in the denominator of the exponential...But the rest is FINALLY correct.