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

The solution for 4y^{''}-4y^'-3y=0,y(0)=5,y^'(0)=0 is y=(5-15/4)e^{(3t)/2}+15/4 e^{-t/2}

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

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

y = (5 - \frac{15}{4}) e^{\frac{3 t}{2}} + \frac{15}{4} e^{- \frac{t}{2}}

Solution steps

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

Solve linear ODE:

y = (5 - \frac{15}{4}) e^{\frac{3 t}{2}} + \frac{15}{4} e^{- \frac{t}{2}}

y = (5 - \frac{15}{4}) e^{\frac{3 t}{2}} + \frac{15}{4} e^{- \frac{t}{2}}

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

y^{^{''}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 1

(D^{2} - 2 D + 5)^{2} y = 0

3 x^{^{''}} + 25 x = 0, x (0) = - 0.5, x^{^{'}} (0) = 2

y^{^{''}} + 6 y^{^{'}} + 8.96 y = 0

What is the solution for 4y^{''}-4y^'-3y=0,y(0)=5,y^'(0)=0 ?
The solution for 4y^{''}-4y^'-3y=0,y(0)=5,y^'(0)=0 is y=(5-15/4)e^{(3t)/2}+15/4 e^{-t/2}