# Thread: First order differential equations

1. ## First order differential equations

Obtain the solution that satisfies the given condition of the differential equation.

(1 + cos^2 x) dy/dx = y(y + 1) sin 2x, given that y = 2 at x = 0

2. Originally Posted by geton
Obtain the solution that satisfies the given condition of the differential equation.

(1 + cos^2 x) dy/dx = y(y + 1) sin 2x, given that y = 2 at x = 0

Separate the variables.

3. But how can I solve this part?

∫ 1/(y)(y+1) dy

4. Originally Posted by geton
But how can I solve this part?

∫ 1/(y)(y+1) dy

Partial fractions.

5. Originally Posted by geton
(1 + cos^2 x) dy/dx = y(y + 1) sin 2x, given that y = 2 at x = 0
$\displaystyle \left( {1 + \cos ^2 x} \right)\,\frac{{dy}} {{dx}} = y(y + 1)\sin 2x \iff \frac{1} {{y(y + 1)}}\,dy = \frac{{\sin 2x}} {{1 + \cos ^2 x}}\,dx.$

Now integrate both sides. (Don't forget integration constant.)

RHS integral is solved here, second one can be easily solved by noting that $\displaystyle (y+1)-y=1,$ form there you just split the original ratio into two fractions.

6. I tried on that way, please check it right or wrong.

After integrating,

1/y – 1/(y+1) = - ln |1 + cos^2 x| + C

Since y =2 at x = 0 this gives
1/2 - 1/3 = - ln 2 + C
C = 1/6 + ln 2

So the solution is,
1/y – 1/(y+1) = - ln |1 + cos^2 x| + 1/6 + ln 2
1/y – 1/(y+1) = ln |2/(1 + cos^2 x)| + 1/6

I did another solution, please check it too.

Can we consider this?
∫ (sin 2x)/(1+cos^2 x)
= ∫2(sin 2x)/(cos 2x + 3)

But still I can’t find my book’s answer, which is (y+1)/y = ¾ (1+cos^2 x)

7. Originally Posted by geton
I tried on that way, please check it right or wrong.

After integrating,

1/y – 1/(y+1) = - ln |1 + cos^2 x| + C
You made a big mistake here

You only integrated the RHS, what about the LHS? It's not integrated

Try again.

(See my signature.)

8. Originally Posted by Krizalid
You made a big mistake here

You only integrated the RHS, what about the LHS? It's not integrated

Try again.

(See my signature.)
Yes that was great mistake.