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.

Let (dy)/(dx) = (y phi'(x)-y^(2))/(phi(x)) , where phi(x) is a function satisfies phi(1) = 1, phi(4) = 1296 . If y(1) = 1 then y(4) is equal to__________

The general solution of the differential equation, y^(prime)+yphi^(prime)(x)-phi(x)phi^(prime)(x)=0 , where phi(x) is a known function, is

if y+x(dy)/(dx)=x(phi(xy))/(phi'(xy)) then phi(xy) is equation to

If phi (x)=phi '(x) and phi(1)=2 , then phi(3) equals

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.

The second derivative of a single valued function parametrically represented by x=phi(t) and y=psi(t) (where phi(t) and psi(t) are different function and phi'(t)ne0) is given by

If y={f(x)}^(phi(x)),"then"(dy)/(dx) is

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

According to Leibritz differentiation under the sign of integration can be performed as as below. (i) (d)/(dx)[int_(phi(x))^(Psi(x))f(t)dt]=f{Psi(x)}xx(d)/(dx){Psi(x)}-f{phi(x)}xx(d)/(dx){phi(x)} (ii) (d)/(dx)[int_(phi(x))^(Psi(x))f(x,t)dt]=int_(phi(x))^(Psi(x))(del)/(delx)(f(x,t)dt)+f(x,Psi(x))xx(d)/(dx)Psi(x)-f(x,phi(x))xx(d)/(dx)(phi(x)) The points of maximum of the function f(x)=int_(0)^(x^(2))(t^(2)-5t+4)/(2+e^(t))dt