If `phi(x)` is a diferential real-valuted function satisfying `phi'(x)+2phi(x)le1.` prove that `phi(x)-(1)/(2)` is a non-incerasing function of x.

If phi(x) is a differential real-valued function satisfying phi'(x)+2phi(x)le1, then the value of 2phi(x) is always less than or equal to …….

If phi (x) is a differentiable real valued function satisfying phi (x) +2phi le 1 , then it can be adjucted as e^(2x)phi(x)+2e^(2x)phi(x)lee^(2x) or (d)/(dx)(e^(2)phi(x)-(e^(2x))/(2))le or (d)/(dx)e^(2x)(phi(x)-(1)/(2))le0 Here e^(2x) is called integrating factor which helps in creating single differential coefficeint as shown above. Answer the following question: If p(1)=0 and dP(x)/(dx)ltP(x) for all xge1 then

Solve (dy)/(dx)+y phi'(x)=phi(x).phi'(x), "where" " "phi(x) is a given function.

Solve (dy)/(dx)=(yphi'(x)-y^(2))/(phi(x)),"where" " "phi(x) is a given function.

If phi(x) is a differentiable function, then the solution of the different equation dy+{y phi' (x) - phi (x) phi'(x)} dx=0, is

If phi(x) is a differentiable function, then the solution of the different equation dy+{y phi' (x) - phi (x) phi'(x)} dx=0, is

Let f(x)=x+1 and phi(x)=x-2. Then the value of x satisfying |f(x)+phi(x)|=|f(x)|+|phi(x)| are :

If phi(x) is a differentiable function AAx in Rand a inR^(+) such that phi(0)=phi(2a),phi(a)=phi(3a)and phi(0)nephi(a), then show that there is at least one root of equation phi'(x+a)=phi'(x)"in"(0,2a).

Let f(x)=x amd g(x)=|x| for all x in R . Then the function phi(x)"satisfying"{phi(x)-f(x)}^(2)+{phi(x)-g(x)}^(2) =0 is

Let f(x) and phi(x) are two continuous function on R satisfying phi(x)=int_(a)^(x)f(t)dt, a!=0 and another continuous function g(x) satisfying g(x+alpha)+g(x)=0AA x epsilonR, alpha gt0 , and int_(b)^(2k)g(t)dt is independent of b If f(x) is an odd function, then