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

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

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

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

y = (1 - \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 6 e ^{\frac{5}{2}} + 2 e}{9 e ^{4} - 1}) e^{\frac{3 t}{2}} + \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 6 e ^{\frac{5}{2}} + 2 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 - 6 e ^{\frac{5}{2}} + 2 e}{9 e ^{4} - 1}) e^{\frac{3 t}{2}} + \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 6 e ^{\frac{5}{2}} + 2 e}{9 e ^{4} - 1} e^{- \frac{t}{2}}

y = (1 - \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 6 e ^{\frac{5}{2}} + 2 e}{9 e ^{4} - 1}) e^{\frac{3 t}{2}} + \frac{9 e ^{4} - 3 e ^{\frac{3}{2}} e - 6 e ^{\frac{5}{2}} + 2 e}{9 e ^{4} - 1} e^{- \frac{t}{2}}

y^{^{''}} + 6 y^{^{'}} + 9 y = 0, y (0) = e^{- 3 x}

2 y^{^{''}} - 7 y^{^{'}} + 5 y = 0, y (0) = 4, y^{^{'}} (0) = 5.5

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

\frac{d ^{2} y}{d t ^{2}} + 18 \frac{d y}{d t} + 81 y = 0

6 y^{^{''}} + 120 y^{^{'}} + 50 y = 0

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