If `x^2+y^2=1` then `(y'=dy/dx, y''=(d^2y)/dx^2)`

If x^(2)+y^(2)=1 then (y'=(dy)/(dx),y''=(d^(2)y)/(dx^(2)))

If x=logp and y=1/p ,then (a) (d^2y)/(dx^2)-2p=0 (b) (d^2y)/(dx^2)+y=0 (c) (d^2y)/(dx^2)+(dy)/(dx)=0 (d) (d^2y)/(dx^2)-(dy)/(dx)=0

If x=logp and y=1/p ,then (a) (d^2y)/(dx^2)-2p=0 (b) (d^2y)/(dx^2)+y=0 (c) (d^2y)/(dx^2)+(dy)/(dx)=0 (d) (d^2y)/(dx^2)-(dy)/(dx)=0

If x=logp and y=1/p ,then (a) (d^2y)/(dx^2)-2p=0 (b) (d^2y)/(dx^2)+y=0 (c) (d^2y)/(dx^2)+(dy)/(dx)=0 (d) (d^2y)/(dx^2)-(dy)/(dx)=0

If x=log pandy=(1)/(p), then (a) (d^(2)y)/(dx^(2))-2p=0 (b) (d^(2)y)/(dx^(2))+y=0 (c) (d^(2)y)/(dx^(2))+(dy)/(dx)=0( d) (d^(2)y)/(dx^(2))-(dy)/(dx)=0

If y= (sin^(-1) x)^2 , then (1-x^2) (d^2 y)/(dx^2) - x (dy/dx) = ..........

If e^y(x+1)=1 show that (d^2y/dx^2) = (dy/dx)^2 ਹੈ।

If e^y(x+1)=1 . Show that (d^2y)/dx^2=(dy/dx)^2

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

Form the differential equation by eliminating A,B from y=e^(5x) (Ax+B) is (A) (d^(2)y)/dx^(2) +10(dy /dx) +25y=0 (B) (d^(2)y) /dx^(2) -10(dy /dx)+25y=0 (C) (d^(2)y) /dx^(2) +10(dy /dx) -25y=0 (D) (d^(2)y) /dx^(2) -10(dy /dx) -25y=0