What is the solution for x^{''}+x^'+0.25x=0,x(0)=2,x^'(0)= 1/3 ?

The solution for x^{''}+x^'+0.25x=0,x(0)=2,x^'(0)= 1/3 is x=2e^{-t/2}+(4e^{-t/2}t)/3

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x^{''}+x^'+0.25x=0,x(0)=2,x^'(0)= 1/3

x^{^{''}} + x^{^{'}} + 0.25 x = 0, x (0) = 2, x^{^{'}} (0) = \frac{1}{3}

x = 2 e^{- \frac{t}{2}} + \frac{4 e ^{- \frac{t}{2}} t}{3}

Solution steps

x^{''} (t) + x^{'} (t) + 0.25 x = 0

Solve linear ODE:

x = 2 e^{- \frac{t}{2}} + \frac{4 e ^{- \frac{t}{2}} t}{3}

x = 2 e^{- \frac{t}{2}} + \frac{4 e ^{- \frac{t}{2}} t}{3}

What is the solution for x^{''}+x^'+0.25x=0,x(0)=2,x^'(0)= 1/3 ?
The solution for x^{''}+x^'+0.25x=0,x(0)=2,x^'(0)= 1/3 is x=2e^{-t/2}+(4e^{-t/2}t)/3