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Popular >

y^'-10y^'+25y=0

Solution

y^{^{'}} - 10 y^{^{'}} + 25 y = 0

Solution

y = e^{\frac{25 t + c _{1}}{9}}

Solution steps

y^{'} (t) - 10 y^{'} (t) + 25 y = 0

Solve separable ODE:

y = e^{\frac{25 t + c _{1}}{9}}

y = e^{\frac{25 t + c _{1}}{9}}

Popular Examples

(dy)/(dx)-0.075y=-9000

\frac{d y}{d x} - 0.075 y = - 9000

y^{^{'}} = e^{y + t}

y*(dy)/(dx)=x*(x-e^x)/(e^{y^2)}

y \cdot \frac{d y}{d x} = x \cdot \frac{x - e ^{x}}{e ^{y^{2}}}

x^2y^'+1/x y= 2/x y^{-3}

x^{2} y^{^{'}} + \frac{1}{x} y = \frac{2}{x} y^{- 3}

(dy)/(dt)=0.07-y/(945)

\frac{d y}{d t} = 0.07 - \frac{y}{945}

Frequently Asked Questions (FAQ)

What is the solution for y^'-10y^'+25y=0 ?
The solution for y^'-10y^'+25y=0 is y=e^{(25t+c_{1})/9}