# Ordinary Differential Equations

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• Nov 16th 2010, 01:37 PM
Anteeks
Ordinary Differential Equations
Need help with following three differential equations. Help is really appreciated!

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

Second:

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

Third:

$\displaystyle \displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)$ that have the value 4 when x=0
• Nov 16th 2010, 03:13 PM
mr fantastic
Quote:

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

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

Second:

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

Third:

$\displaystyle \displaystyle \frac{dy}{dx} = x^5 (3^2+y^2)$ 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.
• Nov 16th 2010, 03:47 PM
Anteeks
Thank you!

First equation:

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

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

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

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

Is this right?

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

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

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

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

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

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

Thats how far I can go.
• Nov 17th 2010, 04:41 AM
HallsofIvy
Quote:

Originally Posted by Anteeks
Thank you!

First equation:

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

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

Quote:

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

Quote:

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

Quote:

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

Quote:

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

Quote:

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

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

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

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

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

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

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