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}

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