Inital value problem

• Mar 5th 2013, 12:20 PM
landmark
Inital value problem
I need a little help as I am not sure of my footing
I have to Solve the initial value problem

dy/dx = cos(5x)/7+sin(5x), y=6 when x=0
• Mar 5th 2013, 12:59 PM
Siron
Re: Inital value problem
Note that the ODE is separable.
• Mar 5th 2013, 01:03 PM
landmark
Re: Inital value problem
I do not understand what you mean
• Mar 5th 2013, 01:11 PM
Siron
Re: Inital value problem
I mean:

$\frac{dy}{dx} = \frac{\cos(5x)}{7}+\sin(5x)\ \mbox{with} \ y=6 \ \mbox{when} \ x=0$
$\Leftrightarrow dy = \left[\frac{\cos(5x)}{7}+\sin(5x)\right]dx$
$\Leftrightarrow \int dy = \int \left[\frac{\cos(5x)}{7}+\sin(5x)\right]dx$

Can you continue?
• Mar 5th 2013, 01:19 PM
landmark
Re: Inital value problem
I can follow that but I think I wrote the initial problem badly it should read (cos⁡(5x))/(7+sin⁡(5x))
• Mar 5th 2013, 03:18 PM
Soroban
Re: Inital value problem
Hello, landmark!

Quote:

$\frac{dy}{dx} \,=\, \frac{\cos(5x)}{7+\sin(5x)} \quad y=6 \text{ when }x=0.$

We have: . $dy \;=\;\frac{\cos(5x)}{7 + \sin(5x)}\,dx$

Integrate: . $\int dy \;=\;\int\frac{\cos(5x)\,dx}{7 + \sin(5x)}$

Let $u \:=\: 7 + \sin(5x) \quad\Rightarrow\quad du \:=\:5\cos(5x)\,dx \quad\Rightarrow\quad \cos(5x)\,dx \:=\:\tfrac{1}{5}du$

Substitute: . $y \:=\:\int\frac{\frac{1}{5}\,du}{u} \:=\:\tfrac{1}{5}\int\frac{du}{u} \:=\;\tfrac{1}{5}\ln |u|+C$

Back-substitute: . $y \:=\:\tfrac{1}{5}\ln|7 + \sin(5x)|+ C$

Initial values: $y = 6,\,x = 0$

. . $6 \;=\;\tfrac{1}{5}\ln|7 + \sin(0)| + C \quad\Rightarrow\quad C \:=\:6 - \tfrac{1}{5}\ln7$

Therefore: . $y \;=\;\tfrac{1}{5}\ln(7 + \sin(5x)| + 6 - \tfrac{1}{5}\ln 7$

. . . . . . . . . $y \;=\;\tfrac{1}{5}\left[\ln|7 + \sin(5x)| - \ln 7\right] + 6$

. . . . . . . . . $y \;=\;\tfrac{1}{5}\ln \left|\frac{7+\sin(5x)}{7}\right| + 6$

. . . . . . . . . $y \;=\;\tfrac{1}{5}\ln\left|1 + \tfrac{1}{7}\sin(5x)\right| + 6$
• Mar 6th 2013, 11:01 AM
landmark
Re: Inital value problem
I was half way there, but was struggling.
Thanks a million
Landmark