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

The solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=3 is y=(1+3/4)e^{(3t)/2}-3/4 e^{-t/2}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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

y^{^{''}} + 5 \cdot y = 0

y^{^{''}} - 4 y + 5 y = 0

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

5 \frac{d ^{2} x}{d t ^{2}} + 3 \frac{d x}{d t} + 6 x = 0

What is the solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=3 ?
The solution for 4y^{''}-4y^'-3y=0,y(0)=1,y^'(0)=3 is y=(1+3/4)e^{(3t)/2}-3/4 e^{-t/2}