What is the solution for x^{''}+1.1875x^'+0.375=0 ?

The solution for x^{''}+1.1875x^'+0.375=0 is x=c_{1}+c_{2}e^{-(2.375t)/2}-0.31578…t

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x^{''}+1.1875x^'+0.375=0

x^{^{''}} + 1.1875 x^{^{'}} + 0.375 = 0

x = c_{1} + c_{2} e^{- \frac{2.375 t}{2}} - 0.31578 \dots t

Solution steps

x^{''} (t) + 1.1875 x^{'} (t) + 0.375 = 0

Solve linear ODE:

x = c_{1} + c_{2} e^{- \frac{2.375 t}{2}} - 0.31578 \dots t

x = c_{1} + c_{2} e^{- \frac{2.375 t}{2}} - 0.31578 \dots t

y^{^{''}} - y^{^{'}} - 2 y = 4, x^{2} y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} + 4 y^{^{'}} + 5 y = 4 + 12 t - 5 t^{2}

y^{^{''}} + 7 y^{^{'}} = 196 sin (7 t) + 588 cos (7 t)

x^{^{''}} + 49 x = 96 cos (5 t)

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

What is the solution for x^{''}+1.1875x^'+0.375=0 ?
The solution for x^{''}+1.1875x^'+0.375=0 is x=c_{1}+c_{2}e^{-(2.375t)/2}-0.31578…t