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

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

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

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

y = 2 e^{\frac{t}{2}} - 4 e^{\frac{t}{2}} t

Solution steps

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

Solve linear ODE:

y = 2 e^{\frac{t}{2}} - 4 e^{\frac{t}{2}} t

y = 2 e^{\frac{t}{2}} - 4 e^{\frac{t}{2}} t

3 y^{^{''}} + 8 y^{^{'}} + 5 y = 0

6 x^{^{''}} + 8 x^{^{'}} + 35 x = 0

3 y^{^{''}} + 2 y^{^{'}} + 2 y = 0

y^{^{''}} + 2 y^{^{'}} + y = 0, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} + y^{^{'}} + \frac{5}{4} y = 0

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