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

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

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

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

y = 2 e^{\frac{t}{2}} - \frac{2 e ^{\frac{t}{2}} t}{3}

Solution steps

y^{''} (t) - y^{'} (t) + 0.25 y = 0

Solve linear ODE:

y = 2 e^{\frac{t}{2}} - \frac{2 e ^{\frac{t}{2}} t}{3}

y = 2 e^{\frac{t}{2}} - \frac{2 e ^{\frac{t}{2}} t}{3}

x^{^{''}} + x^{^{'}} + \frac{1}{2} x = 0, x (0) = 5, x^{^{'}} (0) = 0

y^{^{'''}} - 19 y^{^{''}} + 80 y^{^{'}} + 100 y = 0

y^{^{''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 0

y^{^{''''}} + 3 y^{^{'''}} - 4 y^{^{''}} - 12 y^{^{'}} = 0

8 y^{^{''}} - 8 y^{^{'}} + 100 y = 0

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