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Popular >

f^'(t)=f(t)^{-1}

Solution

f^{^{'}} (t) = f (t)^{- 1}

Solution

f (t) = c_{1} t

Solution steps

f^{'} (t) = f (t) t^{- 1}

Solve separable ODE:

f (t) = c_{1} t

f (t) = c_{1} t

Graph

Popular Examples

y^'=2x^3(1+y^2),y(1)=0

y^{^{'}} = 2 x^{3} (1 + y^{2}), y (1) = 0

(dx)/(dt)=8(x^2+1),x(pi/4)=1

\frac{d x}{d t} = 8 (x^{2} + 1), x (\frac{π}{4}) = 1

(dh)/(dt)=-0.03sqrt(h-19)

\frac{d h}{d t} = - 0.03 h - 19

(dy)/(dx)=8(2-y)

\frac{d y}{d x} = 8 (2 - y)

y^'=4xy-2x,y(0)=4

y^{^{'}} = 4 x y - 2 x, y (0) = 4

Frequently Asked Questions (FAQ)

What is the solution for f^'(t)=f(t)^{-1} ?
The solution for f^'(t)=f(t)^{-1} is f(t)=c_{1}t