Let the function x(t) and y(t) satisfy the differential equations `dy/dt+ax=0, dy/dt+by=0.` If `x(0)=2,y(0)=1 and (x(1))/(y(1))=3/2,` then `x(t)=y(t)` for t =

If y(t) satisfies the differential equation y'(t)+2y(t)=2e^(-2t),y(0)=2 then y(1) equals

If y(t) satisfies the differential equation y'(t)+2y(t)=2e^(-2t),y(0)=2 then y(1) equals

If y(t) satisfies the differential equation y'(t)+2y(t)=2e^(-2t),y(0)=2 then y(1) equals

dx/dt+ax=0, x(0)=2 , y(0)=1 , dy/dt+by=0 3(x(1)=2(y(1) , and find t if x(t)=y(t)

A curve passing through (2,3) and satisfying the differential equation int_0^x ty(t)dt=x^2y(x),(x >0) is

Let x=x(t) and y=y(t) be solutions of the differential equations (dx)/(dt)+ax=0 and (dy)/(dt)+by=0] , respectively, a,b in R .Given that x(0)=2,y(0)=1 and 3y(1)=2x(1) ,the value of "t" ,for which, [x(t)=y(t) ,is :

A curve passing through (2,3) and satisfying the differential equation int_(0)^(x)ty(t)dt=x^(2)y(x),(x>0) is

If y(t) is the solution of the differential equation (t+1)(dy)/(dt)-ty=1 and y(0)=-1 , then the value of y at t=1 is -