What is the solution for 4y^{''}-y^'-3y=0,y(0)=9,y^'(0)=0.25 ?

The solution for 4y^{''}-y^'-3y=0,y(0)=9,y^'(0)=0.25 is y=4e^t+5e^{-(3t)/4}

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4y^{''}-y^'-3y=0,y(0)=9,y^'(0)=0.25

4 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 9, y^{^{'}} (0) = 0.25

y = 4 e^{t} + 5 e^{- \frac{3 t}{4}}

Solution steps

4 y^{''} (t) - y^{'} (t) - 3 y = 0

Solve linear ODE:

y = 4 e^{t} + 5 e^{- \frac{3 t}{4}}

y = 4 e^{t} + 5 e^{- \frac{3 t}{4}}

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (\frac{1}{3}) = 9, y (0) = 3

5 y^{^{''}} + 80 y = 0

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 11, y^{^{'}} (0) = - 49

4000 y^{^{''}} - 9 y^{^{'}} = 0

y^{^{''}} + b y^{^{'}} + 6 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for 4y^{''}-y^'-3y=0,y(0)=9,y^'(0)=0.25 ?
The solution for 4y^{''}-y^'-3y=0,y(0)=9,y^'(0)=0.25 is y=4e^t+5e^{-(3t)/4}