What is the solution for 4y^{''}-y^'-3y=0,y(0)=5,y^'(0)=0 ?

The solution for 4y^{''}-y^'-3y=0,y(0)=5,y^'(0)=0 is y=(5-20/7)e^t+20/7 e^{-(3t)/4}

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4y^{''}-y^'-3y=0,y(0)=5,y^'(0)=0

4 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 5, y^{^{'}} (0) = 0

y = (5 - \frac{20}{7}) e^{t} + \frac{20}{7} e^{- \frac{3 t}{4}}

Solution steps

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

Solve linear ODE:

y = (5 - \frac{20}{7}) e^{t} + \frac{20}{7} e^{- \frac{3 t}{4}}

y = (5 - \frac{20}{7}) e^{t} + \frac{20}{7} e^{- \frac{3 t}{4}}

y^{^{''''}} - 7 y^{^{'''}} + 12 y^{^{''}} = 0

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

y^{^{'''}} + 3 y^{^{''}} - 54 y^{^{'}} = 0

\frac{d ^{2} y}{d x ^{2}} + 2 \frac{d y}{d x} + y = 0

5 y^{^{''}} - 6 y^{^{'}} + y = 0, y (0) = 5, y^{^{'}} (0) = \frac{1}{5}

What is the solution for 4y^{''}-y^'-3y=0,y(0)=5,y^'(0)=0 ?
The solution for 4y^{''}-y^'-3y=0,y(0)=5,y^'(0)=0 is y=(5-20/7)e^t+20/7 e^{-(3t)/4}