1. ## Ordinary Differential Equations

Need help with following three differential equations. Help is really appreciated!

First:
Solve the differential equation $y' - 2 y = 3$, y(0) = 1.

Second:

$\displaystyle \frac{dy}{dx} + 8 \sin(x) y = (x+3)^4 e^{8 \cos(x)}$, y(0)=5.

Third:

$\displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)$ that have the value 4 when x=0

2. Originally Posted by Anteeks
Need help with following three differential equations. Help is really appreciated!

First:
Solve the differential equation $y' - 2 y = 3$, y(0) = 1.

Second:

$\displaystyle \frac{dy}{dx} + 8 \sin(x) y = (x+3)^4 e^{8 \cos(x)}$, y(0)=5.

Third:

$\displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)$ that have the value 4 when x=0
The first and third are seperable. The second uses the integrating factor.

If you need more help, please show all your work and say where you're stuck.

3. Thank you!

First equation:

$e^int{(-2x)dx = e^{(-2x)$

$\displaystyle \frac{dy}{dx}*e^{(-2x))$ $-2y*e^{(-2x)$ $= 3*e^{(-2x)$
-->
$D(y*e^{(-2x))$ $=3*e^{(-2x)$
-->
$y*e^{(-2x)$= $int(3*e^{(-2x))dy$ $=-6*e^{(-2x)$ $+ C$

$y=-6+C*e^{(2x)$
$y(0)=6$ $--> -6+C*e^0=6, C=11$

Which gives $y=-6+11*e^{(2x)$

Is this right?

Second:
integrating factor = $e^{8 \cos(x)}$

$\displaystyle \frac{dy}{dx} *e^{8 \cos(x$) $+ 8 \sin(x) *$ $e^{8 \cos(x)$ $y = (x+3)^4$ $e^{8 \cos(x)}*$ $e^{8 \cos(x)$

Thats about how far I can go, and don't know if its right.

Third:
$\displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)$

Divide with $(3^2+y^2)$ on both rows and integrate both:

$\displaystyle \frac{dy}{dx}/(3^2+y^2) = x^5$

Thats how far I can go.

4. Originally Posted by Anteeks
Thank you!

First equation:

$e^int{(-2x)}dx = e^{(-2x)$
Yes, that first equation is linear and so you can get an integrating factor this way. It is $\int -2 dx= -2x$, not $\int -2x dx$. After that you are good

$\displaystyle \frac{dy}{dx}*e^{(-2x))$ $-2y*e^{(-2x)$ $= 3*e^{(-2x)$
-->
$D(y*e^{(-2x))$ $=3*e^{(-2x)$
-->
$y*e^{(-2x)$= $int(3*e^{(-2x))dy$ $=-6*e^{(-2x)$ $+ C$
until here. $3\int e^{-2x}dx= -\frac{3}{2}e^{-2x}+ C$, not $-6 e^{-2x}$.

$y=-6+C*e^{(2x)$
$y(0)=6$ $--> -6+C*e^0=6, C=11$
But mr fantastic told you the equation was separable. It is much easier to do $\frac{dy}{dx}= 2y+ 3$ so that $\frac{dy}{2y+ 3}= \frac{dx}$. Integrating, $ln(2y+ 3)= x+ C$ so that $2y+ 3= ce^{x}$.

Which gives $y=-6+11*e^{(2x)$
Is this right?
Well, it is not hard to check yourself: if $y= -6+ 11e^{2x}$ then $y'= -22e^{-2x}$ while $-2y= 12- 22e^{-2x}$. So $y'- 2y= -22e^{-2x}+ 12- 22e^{-2x}= 12- 44e^{-2x}$. No, that is not "3" and so your solution does not check.

Second:
integrating factor = $e^{8 \cos(x)}$
No, the integral of 8 sin(x) is -8 cos(x).

$\displaystyle \frac{dy}{dx} *e^{8 \cos(x$) $+ 8 \sin(x) *$ $e^{8 \cos(x)$ $y = (x+3)^4$ $e^{8 \cos(x)}*$ $e^{8 \cos(x)$

Thats about how far I can go, and don't know if its right.

Third:
$\displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)$

Divide with $(3^2+y^2)$ on both rows and integrate both:

$\displaystyle \frac{dy}{dx}/(3^2+y^2) = x^5$

Thats how far I can go.
One of the things you should have learned long ago is that $\int \frac{du}{1^2+ u^2}= arctan(u)+ C$.

$\frac{1}{3^2+ y^2}= \frac{1}{3^2}\frac{1}{1+ \left(\frac{y}{3}\right)^2}$. Let u= y/3.