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Popular >

y^{''}-2y^'+5=0

Solution

y^{^{''}} - 2 y^{^{'}} + 5 = 0

Solution

y = c_{1} e^{2 t} + c_{2} + \frac{5 t}{2}

Solution steps

y^{''} (t) - 2 y^{'} (t) + 5 = 0

Solve linear ODE:

y = c_{1} e^{2 t} + c_{2} + \frac{5 t}{2}

y = c_{1} e^{2 t} + c_{2} + \frac{5 t}{2}

Graph

Popular Examples

y^{''}-8y^'+20y=200x^2-39

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 39

(d^2y)/(dx^2)-5(dy)/(dx)+6y=e^{4x}

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 6 y = e^{4 x}

y^{''}-2y^'-3y=4

y^{^{''}} - 2 y^{^{'}} - 3 y = 4

y^{''}+2y^'-8y=3e^{-2x}-e^{-x}

y^{^{''}} + 2 y^{^{'}} - 8 y = 3 e^{- 2 x} - e^{- x}

y^{''}-a^2y=sec(ax)

y^{^{''}} - a^{2} y = sec (a x)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}-2y^'+5=0 ?
The solution for y^{''}-2y^'+5=0 is y=c_{1}e^{2t}+c_{2}+(5t)/2