Solving dy/dx = (x-4)(y+1) with y(0)=4

[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.

 

[dextercioby]

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

You may want to check again.

Daniel.

 

[HallsofIvy]

In other words, you factored wrong

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

 

[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.

 

[p53ud0 dr34m5]

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

try that one?

 

[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.

 

[p53ud0 dr34m5]

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

 

[dextercioby]

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

Advice:take a well diserved break...

Daniel.

 

[p53ud0 dr34m5]

oh my dear! i can't believe i missed that! i still must be fried from the weed.