The solution of the differential equation `2x^2y(dy)/(dx)=tan(x^2y^2)-2x y^2,` given `x=1, y=pi/2,` is

Options

A. `sin^(2)y^(2)=e^(x+1)`
B. `sin(x^(2)y^(2))=x`
C. `cosx^(2)y^(2)+x=0`
D. `sin(x^(2)y^(2))=e^(x-1)`

Answer

D

Explanation

`2x^(2)y(dy)/(dx) = tan(x^(2)y^(2))-2xy^(2)`
or `x^(2)2y(dy)/(dx) + y^(2)2x=tan(x^(2)y^(2))`
or `d/(dx)(x^(2)y^(2))=tan(x^(2)y^(2))`
`intcot(x^(2)y^(2))d(x^(2)y^(2))=intdx`
`log(sinx^(2)y^(2))d(x^(2)y^(2))=int(dx)`
or `logsin(x^(2)y^(2))=x+c`
When `x=1, y=sqrt(pi)/2`, then `c=-1`.
Thus, equation of curve is
`logsin(x^(2)y^(2))=x-1`
or `sin(x^(2)y^(2))=e^(x-1)`