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Popular >

ln(y^')=y

Solution

ln (y^{^{'}}) = y

Solution

y = - ln (- t - c_{1})

Solution steps

ln (y^{'} (t)) = y

Solve separable ODE:

y = - ln (- t - c_{1})

y = - ln (- t - c_{1})

Popular Examples

y y^{^{'}} - x y^{2} + x = 0

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

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

e^{-y}sin(x)-(dy)/(dx)cos^2(x)=0

e^{- y} sin (x) - \frac{d y}{d x} cos^{2} (x) = 0

(dy)/(dx)=(17xy)/((ln(y))^6)

\frac{d y}{d x} = \frac{17 x y}{( ln ( y ) ) ^{6}}

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

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

Frequently Asked Questions (FAQ)

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