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Popular >

y^{''}+2y^'+24=0

Solution

y^{^{''}} + 2 y^{^{'}} + 24 = 0

Solution

y = c_{1} + c_{2} e^{- 2 t} - 12 t

Solution steps

y^{''} (t) + 2 y^{'} (t) + 24 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 2 t} - 12 t

y = c_{1} + c_{2} e^{- 2 t} - 12 t

Popular Examples

y^{''}+y=x^2+1,y(0)=2,y(1)=0

y^{^{''}} + y = x^{2} + 1, y (0) = 2, y (1) = 0

(d^2y)/(dx^2)-4(dy)/(dx)+4y=3x-4

\frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + 4 y = 3 x - 4

(d^2x)/(dt^2)+9x=e^t+sin(t)

\frac{d ^{2} x}{d t ^{2}} + 9 x = e^{t} + sin (t)

y^{''}-4y^'+4y=t^2e^{2t}+e^{2t}

y^{^{''}} - 4 y^{^{'}} + 4 y = t^{2} e^{2 t} + e^{2 t}

y^{''}-y=(1-e^2x)

y^{^{''}} - y = (1 - e^{2} x)

Frequently Asked Questions (FAQ)

What is the solution for y^{''}+2y^'+24=0 ?
The solution for y^{''}+2y^'+24=0 is y=c_{1}+c_{2}e^{-2t}-12t