What is the solution for 4y^{''}-9y=0,y(0)=0,y^'(0)=3 ?

The solution for 4y^{''}-9y=0,y(0)=0,y^'(0)=3 is y=e^{(3t)/2}-e^{-(3t)/2}

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4y^{''}-9y=0,y(0)=0,y^'(0)=3

4 y^{^{''}} - 9 y = 0, y (0) = 0, y^{^{'}} (0) = 3

y = e^{\frac{3 t}{2}} - e^{- \frac{3 t}{2}}

Solution steps

4 y^{''} (t) - 9 y = 0

Solve linear ODE:

y = e^{\frac{3 t}{2}} - e^{- \frac{3 t}{2}}

y = e^{\frac{3 t}{2}} - e^{- \frac{3 t}{2}}

y^{^{'''}} - 2 y^{^{''}} - 21 y^{^{'}} - 18 y = 0

4 x^{^{''}} + 8 x^{^{'}} + 4 x = 0

\frac{d ^{4} y}{d x ^{4}} - 16 y = 0

y^{^{''}} - 6 y^{^{'}} = 0, y (0) = 4 - 4 e^{6}, y (1) = 0

y^{^{'''}} - y^{^{''}} - 12 y^{^{'}} = 0

What is the solution for 4y^{''}-9y=0,y(0)=0,y^'(0)=3 ?
The solution for 4y^{''}-9y=0,y(0)=0,y^'(0)=3 is y=e^{(3t)/2}-e^{-(3t)/2}