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Popular >

y-14y^'+50y=0

Solution

y - 14 y^{^{'}} + 50 y = 0

Solution

y = e^{\frac{51 t + c _{1}}{14}}

Solution steps

y - 14 y^{'} (t) + 50 y = 0

Solve separable ODE:

y = e^{\frac{51 t + c _{1}}{14}}

y = e^{\frac{51 t + c _{1}}{14}}

Popular Examples

(dy)/(dx)y=ln^8(sqrt((4+x^2)/(4-x^2)))

\frac{d y}{d x} y = ln^{8} (\frac{4 + x ^{2}}{4 - x ^{2}})

(dy)/y =(4dx)/x

\frac{d y}{y} = \frac{4 d x}{x}

y(dy)/(dx)= x/((x+1))

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

(dy)/(dx)=-4xy^2,y(0)=1

\frac{d y}{d x} = - 4 x y^{2}, y (0) = 1

(dP)/(dt)=-e^{-t}P,P(0)=100

\frac{d P}{d t} = - e^{- t} P, P (0) = 100

Frequently Asked Questions (FAQ)

What is the solution for y-14y^'+50y=0 ?
The solution for y-14y^'+50y=0 is y=e^{(51t+c_{1})/(14)}