What is the solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(1)=5 ?

The solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(1)=5 is

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4y^{''}-4y^'-3y=0,y(0)=1,y^'(1)=5

4 y^{^{''}} - 4 y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (1) = 5

y = (1 - \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 30 e ^{\frac{5}{2}} + 10 e}{9 e ^{4} - 1}) e^{\frac{3 t}{2}} + \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 30 e ^{\frac{5}{2}} + 10 e}{9 e ^{4} - 1} e^{- \frac{t}{2}}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 30 e ^{\frac{5}{2}} + 10 e}{9 e ^{4} - 1}) e^{\frac{3 t}{2}} + \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 30 e ^{\frac{5}{2}} + 10 e}{9 e ^{4} - 1} e^{- \frac{t}{2}}

y = (1 - \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 30 e ^{\frac{5}{2}} + 10 e}{9 e ^{4} - 1}) e^{\frac{3 t}{2}} + \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 30 e ^{\frac{5}{2}} + 10 e}{9 e ^{4} - 1} e^{- \frac{t}{2}}

2 y^{^{''}} - 2 y^{^{'}} + y = 0, y^{0} = - 1, y^{^{'}} 0 = 0

y^{^{''}} + 2 y^{^{'}} - 6 y = 0

\frac{d ^{4} y}{d x ^{4}} + 2 y^{^{'''}} - 2 y^{^{'}} - y = 0

y^{^{''}} - 6 y^{^{'}} + 13 y = 0

y^{^{''}} + 8 y^{^{'}} + 25 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(1)=5 ?
The solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(1)=5 is