# Can someone help me solve this differential equation

#### Reelacmpj

##### New member

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

#### Subhotosh Khan

##### Super Moderator
Staff member
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.
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.....

#### Reelacmpj

##### New member
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
I have done that but I’m talking more towards the end when I have a y= 2/sqrt(1+Ce^8cosx)

#### Reelacmpj

##### New member
This is what I got but the next question asks for y(pi/2)=0 and the solution seems non-trivial

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#### topsquark

##### Full Member
$$\displaystyle cos \left ( \dfrac{ \pi }{2} \right ) = 0$$

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