Question

If y(x) satisfies the differential equation y'-ytanx=2xsecx and y(0)=0 then?

Answer 1

A) y(pi/4)=pi^2/8 sqrt(2)
B)y'(pi/4)=pi^2/18
C)y(pi/3)=pi^2/9
D)y'(pi/3)=4pi/3+2pi^2/3sqrt(3)

Answer 2

Attempt:
I guess its a linear differential equation of type
\[\LARGE \frac{dy}{dx}+PY=Q\]
Where,
P= -tanx
And integrating factor,
\[\LARGE I.F=e^{\int\limits pdx}=cosx\]
After solving a bit,:
\[\LARGE Y \cos x=x^2+c\]
According to given conditions,\(\large c=y\)
\[\LARGE Y(cosx-1)=x^2\]
\[\LARGE y=\frac{x^2}{cosx-1}\]
This is my equation,which doesn't prove any option right..

Answer 3

The "after solving a bit" part might be where the problem is.

Answer 4

c=0 not y.