- Thread starter Reelacmpj
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- Jun 18, 2007

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I'll start you off. I'll skip some steps and expect you to fill in:View attachment 22326

I have been trying for hours to get a general solution but nothing I try is valid. Can anyone help me please.

y'/[y(4-y^2] = sin(x)

dy/[y(2-y)(2+y)] = sin(x)dx

dy[A/y + B/(2-y) + C/(2+y)] = sin(x) dx

Using method of partial fractions - solve for A, B & C and continue.....

I have done that but I’m talking more towards the end when I have a y= 2/sqrt(1+Ce^8cosx)I'll start you off. I'll skip some steps and expect you to fill in:

y'/[y(4-y^2] = sin(x)

dy/[y(2-y)(2+y)] = sin(x)dx

dy[A/y + B/(2-y) + C/(2+y)] = sin(x) dx

Using method of partial fractions - solve for A, B & C and continue.....

View attachment 22329

- Joined
- Aug 27, 2012

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- 959

Thus

\(\displaystyle y \left ( \dfrac{ \pi }{2} \right ) = 0 = \pm \dfrac{2}{ 1 - C e^{ 8 \cdot 0 } } = \pm \dfrac{2}{1 - C}\)

Is there any C that will do this?

What does that mean about the point \(\displaystyle (x, y) = \left ( \dfrac{ \pi }{2} , 0 \right )\)? (Pick a few C values and graph the function and you'll see the problem.)

-Dan