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Popular >

y^'-10y+25y=0

Solution

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

Solution

y = e^{- 15 t + c_{1}}

Solution steps

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

Solve separable ODE:

y = e^{- 15 t + c_{1}}

y = e^{- 15 t + c_{1}}

Graph

Popular Examples

(dy)/(dx)= y/(x(1-x^2))

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

y^'=5e^{x-y},y(0)=ln(7)

y^{^{'}} = 5 e^{x - y}, y (0) = ln (7)

y^{^{'}} + (8 x - 1) y = 0

(dh)/(dt)=(-3/100 sqrt(h-19))

\frac{d h}{d t} = (- \frac{3}{100} h - 19)

(ydy)/(1+y^2)=(xdx)/(2+x^2)

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

Frequently Asked Questions (FAQ)

What is the solution for y^'-10y+25y=0 ?
The solution for y^'-10y+25y=0 is y=e^{-15t+c_{1}}