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Popular Calculus >

y^'*ln(y)-y=0

Solution

y^{^{'}} \cdot ln (y) - y = 0

Solution

y = e^{2 t + c_{1}}, y = e^{- 2 t + c_{1}}

Solution steps

y^{'} (t) \cdot ln (y) - y = 0

Solve separable ODE:

y = e^{2 t + c_{1}}, y = e^{- 2 t + c_{1}}

y = e^{2 t + c_{1}}, y = e^{- 2 t + c_{1}}

Graph

Popular Examples

y^'=((x^2-5))/(y^2),y(0)=3

(dy)/(dx)=0.04-(3y)/(880)

(dy)/(dx)y=12(3x+1)^{2/3}(4x-2)^{5/4}

Frequently Asked Questions (FAQ)

What is the solution for y^'*ln(y)-y=0 ?
The solution for y^'*ln(y)-y=0 is y=e^{sqrt(2t+c_{1)}},y=e^{-sqrt(2t+c_{1)}}