1. ## first-order differential eq.

Solve the inital value problem for y(x):

$xy' + 8y = 3x^4$

with the initial condition: $y(1) = 4$
integrating factor would be $e^{8x}$

$\frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)$

$e^{8x}y = \int e(3x^4)^{8x}$

not sure if what im doing so far is right, having a hard time finishing it.

2. Originally Posted by viet
integrating factor would be $e^{8x}$

$\frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)$

$e^{8x}y = \int e(3x^4)^{8x}$

not sure if what im doing so far is right, having a hard time finishing it.
The integrating factor is $x^7$

multiply through by x^7

$\Rightarrow x^8 y' + 8x^7 y = 3x^{11}$

$\Rightarrow \left( x^8 y \right)^{ \prime} = 3x^{11}$

Now continue

3. Originally Posted by viet
integrating factor would be $e^{8x}$

$\frac{d}{dx} (y* e^{8x}) = e^{8x}(3x^4)$

$e^{8x}y = \int e(3x^4)^{8x}$

not sure if what im doing so far is right, having a hard time finishing it.
i will recheck this. i vaguely remember some method where you always use e^(some power) for the integrating factor, but i'm not sure if you did it correctly (i rarely remember formulas or routines verbatim, i just know how to work stuff out). i know the last line is wrong though, since 8x is the power of e, not 3x^4.

if you want to know how i got my integrating factor, ask

4. $xy'+8y=3x^4 \mbox{ on }(0,\infty)$

Then,

$y' + \frac{8}{x} \cdot y = 3x^3$

The solution is given by,

$y = \frac{1}{I(x)}\cdot \left( \int I(x)Q(x) dx +C \right)$

Where $I(x) = \exp \left( \int \frac{8}{x} dx \right)$ and $Q(x) = 3x^3$.

5. ok i corrected this.

the integrating factor i got is $x^8$ i got it by:
$xy' +8y = 3x^4$

$y' + \frac{8}{x}y = 3x^3$

$e^{{\int}\frac{8}{x}dx} = x^8$

multiply both sides by $x^8$
$x^9y' + 8x^8y = 3x^12$

$\frac{d}{dx}(x^9y) = 3x^12$

$x^9y = \int{3x^{12}}dx$

$x^9y = \frac{3x^{13}}{13}+C$

$y = (\frac{3x^{13}}{13}+C)x^{-9}$

$y = \frac{3x^{4}}{13}+Cx^{-9}$

i guess now i just plug in y=1, x=4. to solve for C

6. Originally Posted by viet
ok i corrected this.

the integrating factor i got is $x^8$ i got it by:
$xy' +8y = 3x^4$

$y' + \frac{8}{x}y = 3x^3$

$e^{{\int}\frac{8}{x}dx} = x^8$

multiply both sides by $x^8$
$x^9y' + 8x^8y = 3x^12$

$\frac{d}{dx}(x^9y) = 3x^12$

$x^9y = \int{3x^{12}}dx$

$x^9y = \frac{3x^{13}}{13}+C$

$y = (\frac{3x^{13}}{13}+C)x^{-9}$

$y = \frac{3x^{4}}{13}+Cx^{-9}$

i guess now i just plug in y=1, x=4. to solve for C
when you multiply through by x^8 you should get what i got. don't multiply the original equation. multiply the one you simplified to when you dived through by x

7. Originally Posted by Jhevon
when you multiply through by x^8 you should get what i got. don't multiply the original equation. multiply the one you simplified to when you dived through by x
i continue on from what you got, and got:

$y = \frac{1}{4}x^4 + C{x^{-8}}$

8. Originally Posted by viet
i continue on from what you got, and got:

$y = \frac{1}{4}x^4 + C{x^{-8}}$
ok, so far. but don't forget the initial condition, they expect you to find C

9. $1 = 64 + C(4^{-8})$

$-63 = C(\frac{1}{65536})$

$-4128768 = C$

i think i did it wrong

10. Originally Posted by viet
$1 = 64 + C(4^{-8})$

$-63 = C(\frac{1}{65536})$

$-4128768 = C$

i think i did it wrong
yes you did

it says $y(1) = 4$

so $y(1) = 4 = \frac {1}{4} (1)^4 + C(1)^{-8}$

$\Rightarrow 4 = \frac {1}{4} + C$

Now continue

11. thanks i feel pretty dumb.
final answer is $y = \frac{1}{4}x^4 + \frac{15}{4}{x^{-8}}$