The solution of the differentia equation `(dy)/(dx)+(2y)/x=0` with `y(1)=1` is given by `y=1/(x^2)` b. `x=1/(y^2)` c. `x=1/y` d. `y=1/x`

Options

A. a+b+c
B. `y(pi)=-3/5`
C. y(x) is decreasing in `(-pi/2,0)`
D. maximum value of y(x) is `1/sqrt5`

Answer

D

Explanation

It is a linear equation of the form `(dy)/(dx)+py=Q`
Where p=2, Q= cos x
Then I.F. = `e^(2x)`
Hence , the general solution is `y.e^(2x)=int e^(2x)cos x dx +c`
`y.e^(2x)=e^(2x)/5` (2 cos x + sin x ) +c
But, given that y(0)=`2/5 rArr ` c=0
So `y.e^(2x)=e^(2x)/5(2 cos x = sin x) rArr y=2/5 cos x + 1/5 sin x`
Now `y(pi/2)=1/5, y(pi)=(-2)/5, (dy)/(dx) > 0 AA x in (-pi/2, 0)`
So y(x) is increasing in `((-pi)/2,0)`
And maximum value of y(x) is `1/sqrt5`