" (iv) "y=Ax" and "xy'=y,(x!=0)" .(NCERT) "

y = Ax : xy' = y (x ne 0)

y = Ax : xy' = y (x ne 0)

y = Ax : xy' = y (x ne 0)

Verify that the given function is a solution of the corresponding differential equation: y=Ax: xy\'=y (x!=0)

If log(x+y)=2xy then y'(0)=

If log(x+y)=2xy, then y'(0) is

If log(x+y)=2xy, then y'(0) is

If log (x+y) = 2xy, then y^(')(0) =

If log(x+y)=2xy, then y'(0) is

Verify the differential equation y = Ax : xy' = y (x ne 0)