`(dy)/(dx)tan x=y+1,y(2)=0`

Solve (dy)/(dx)+y tan x=y^(2)sec x

Solve (dy)/(dx)+y tan x=y^(2)sec x

Solve (dy)/(dx)+y tan x=y^(2)sec x

Solve each of the following initial value problem: (dy)/(dx)+y tan x=2x+x^(2)tan x,y(0)=1

if y(x) is satisfy the differential equation (dy)/(dx)=(tan x-y)sec^(2)x and y(0)=0. Then y=(-(pi)/(4)) is equal to

(dy)/(dx) + y tan x = 2x + x ^(2) tan x , given that y = 1 when x=0

(dy) / (dx) = tan ^ (2) (x + y)

For each of the following initial value problems verify that the accompanying functions is a solution. (i) x(dy)/(dx)=1, y(1)=0 => y=logx (ii) (dy)/(dx)=y , y(0)=1 => y=e^x (iii) (d^2y)/(dx^2)+y=0, y(0)=0, y^(prime)(0)=1 => y=sinx (iv) (d^2y)/(dx^2)-(dy)/(dx)=0, y(0)=2, y^(prime)(0)=1 => y=e^x+1 (v) (dy)/(dx)+y=2, y(0)=3 => y=e^(-x)+2

If y=sin(sin x), prove that (d^(2)y)/(dx^(2))+tan x(dy)/(dx)+y cos^(2)x=0

If (dy)/(dx)+y tan x=sin2x and y(0)=1, then y(pi) is equal to: