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
Please help me to solve that.
$\displaystyle \left( {1 + \cos ^2 x} \right)\,\frac{{dy}}
{{dx}} = y(y + 1)\sin 2x \iff \frac{1}
{{y(y + 1)}}\,dy = \frac{{\sin 2x}}
{{1 + \cos ^2 x}}\,dx.$
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.
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)