# Last Diff Equation for today

• Apr 30th 2006, 11:41 AM
Natasha1
Last Diff Equation for today
I am asked to find the general solution, using the method of separating the variables of the following DE

dy/dx = (2xcos x)/y where y > 0

Using the method of seperating the variables I get:

int y.dy = int 2xcosx dx

which basically comes to (after a little working)

= 2 (cos x + xsinx)

but where do I go from here to get an integration constant?

2) If y = 2 when x = 0, find y in terms of x

Could someone help me on this one

3) Explain why your answer may not be used for x=pi. Comment in relation to the solution curve through (0,2).
• Apr 30th 2006, 11:45 AM
ThePerfectHacker
Quote:

Originally Posted by Natasha1
I am asked to find the general solution, using the method of separating the variables of the following DE

dy/dx = (2xcos x)/y where y > 0

Using the method of seperating the variables I get:

int y.dy = int 2xcosx dx

which basically comes to (after a little working)

= 2 (cos x + xsinx)

but where do I go from here to get a integration constant?

2) If y = 2 when x = 0, find y in terms of x

Could someone help me on this one

3) Explain why your answer may not be used for x=pi. Comment in relation to the solution curve through (0,2).

Thus,
$\frac{1}{2}y^2=2(\cos x+x\sin x)+C$
But $(x,y)=(0,2)$
Thus,
$2=2(1+0)+C$
Thus,
$C=0$
Which gives,
$\frac{1}{2}y^2=2\cos x+2x\sin x$
Thus,
$y=2\sqrt{ \cos x+x\sin x}$

You cannot use $x=\pi$ because you get a negative in the radical, which brings non-real numbers.
• Apr 30th 2006, 11:58 AM
Natasha1
Quote:

Originally Posted by ThePerfectHacker
Thus,
$\frac{1}{2}y^2=2(\cos x+x\sin x)+C$
But $(x,y)=(0,2)$
Thus,
$2=2(1+0)+C$
Thus,
$C=0$
Which gives,
$\frac{1}{2}y^2=2\cos x+2x\sin x$
Thus,
$y=2\sqrt{ \cos x+x\sin x}$

You cannot use $x=\pi$ because you get a negative in the radical, which brings non-real numbers.

What about the relation to the solution curve through (0,2)?
• Apr 30th 2006, 12:14 PM
CaptainBlack
Quote:

Originally Posted by Natasha1
What about the relation to the solution curve through (0,2)?

PH found the general soltion, then substituted the initial condition
into the general solution to obtain his final solution.

RonL
• Apr 30th 2006, 12:21 PM
Natasha1
Quote:

Originally Posted by CaptainBlack
PH found the general soltion, then substituted the initial condition
into the general solution to obtain his final solution.

RonL

If the general solution is y = 2 SQU ROOT OF cosx + xsinsx

and subsitution the initial condition y = 2 and x = 0

we get

2 = 2 SQU ROOT OF cos 0 + 0sin0
2=2

?
• Apr 30th 2006, 12:53 PM
CaptainBlack
Quote:

Originally Posted by Natasha1
If the general solution is y = 2 SQU ROOT OF cosx + xsinsx

and subsitution the initial condition y = 2 and x = 0

we get

2 = 2 SQU ROOT OF cos 0 + 0sin0
2=2

?

Without checking in detail, my reading of PH's posting is that the general
solution is:

$
\frac{1}{2}y^2=2(\cos x+x\sin x)+C
$
,

into which he substitutes $y=2$ when $x=0$, to get

$
\frac{1}{2}y^2=2\cos x+2x\sin x
$

or:

$
y=2\sqrt{ \cos x+x\sin x}
$

RonL