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

The solution for 4y^{''}-4y^'+(-3)y=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^'+(-3)y=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^{^{''}} + 2 y^{^{'}} - y = 0

y^{^{''}} + 0.25 y = 0

\frac{17}{160} y^{^{''}} + \frac{4}{10} y^{^{'}} + 2 y = 0

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

x^{^{''}} - 9 x^{^{'}} - 14 x = 0

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