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Popular >

x^'=e^{-x+c}

Solution

x^{^{'}} = e^{- x + c}

Solution

x = ln (e^{c} t + c_{1})

Solution steps

x^{'} (t) = e^{- x + c}

Solve separable ODE:

x = ln (e^{c} t + c_{1})

x = ln (e^{c} t + c_{1})

Graph

Popular Examples

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

\frac{d y}{d x} = y^{2} x + y^{2} cos (x)

y^'=((1+x^2))/y

y^{^{'}} = \frac{( 1 + x ^{2} )}{y}

(dy)/(dx)= x/(y^8)

\frac{d y}{d x} = \frac{x}{y ^{8}}

5x^4dx+20y^{19}dy=0

5 x^{4} d x + 20 y^{19} d y = 0

(dP)/(dt)=0.06P(1-P/(350))

\frac{d P}{d t} = 0.06 P (1 - \frac{P}{350})

Frequently Asked Questions (FAQ)

What is the solution for x^'=e^{-x+c} ?
The solution for x^'=e^{-x+c} is x=ln(e^ct+c_{1})