What is the solution for ax^{''}+bx^'+c=0 ?

The solution for ax^{''}+bx^'+c=0 is x=c_{1}+c_{2}e^{-(bt)/a}-(ct)/b

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ax^{''}+bx^'+c=0

a x^{^{''}} + b x^{^{'}} + c = 0

x = c_{1} + c_{2} e^{- \frac{b t}{a}} - \frac{c t}{b}

Solution steps

a x^{''} (t) + b x^{'} (t) + c = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- \frac{b t}{a}} - \frac{c t}{b}

x = c_{1} + c_{2} e^{- \frac{b t}{a}} - \frac{c t}{b}

\frac{d ^{2} y}{d t ^{2}} - 3 \frac{d y}{d t} - 10 y = 4 e^{5 t}

y^{^{''}} - y^{^{'}} - 12 y = e^{2} x + e^{x} - 2

y^{^{''}} + 6 y^{^{'}} + 5 y = 36 e^{x}, y (0) = - 2, y^{^{'}} (0) = 4

y^{^{''}} + y^{^{'}} - 12 y = \frac{e ^{x}}{x}

(6 D^{2} + 4 D + 2) y = x e^{- 2 x}

What is the solution for ax^{''}+bx^'+c=0 ?
The solution for ax^{''}+bx^'+c=0 is x=c_{1}+c_{2}e^{-(bt)/a}-(ct)/b