What is the solution for (x^{''})/2+10x^'+100x=150 ?

The solution for (x^{''})/2+10x^'+100x=150 is x=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2

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(x^{''})/2+10x^'+100x=150

\frac{x ^{^{''}}}{2} + 10 x^{^{'}} + 100 x = 150

x = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

Solution steps

\frac{x ^{''} ( t )}{2} + 10 x^{'} (t) + 100 x = 150

Solve linear ODE:

x = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

x = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

y^{^{''}} + a y^{^{'}} + b y = t + e^{3 t}

\frac{d ^{2} y}{d t ^{2}} - 3 \frac{d y}{d t} - 40 y = 5 + 4 t

y^{^{''}} + 4 x^{2} + 4 y = 6

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} + 8 y = x e^{x}

y^{^{''}} - 9 y^{^{'}} + 14 y = 12 - e^{x}

What is the solution for (x^{''})/2+10x^'+100x=150 ?
The solution for (x^{''})/2+10x^'+100x=150 is x=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2