What is the solution for 36y^{''}-81y=0,y(0)=0,y^'(0)=1 ?

The solution for 36y^{''}-81y=0,y(0)=0,y^'(0)=1 is y= 1/3 e^{(3t)/2}-1/3 e^{-(3t)/2}

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36y^{''}-81y=0,y(0)=0,y^'(0)=1

36 y^{^{''}} - 81 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y = \frac{1}{3} e^{\frac{3 t}{2}} - \frac{1}{3} e^{- \frac{3 t}{2}}

Solution steps

36 y^{''} (t) - 81 y = 0

Solve linear ODE:

y = \frac{1}{3} e^{\frac{3 t}{2}} - \frac{1}{3} e^{- \frac{3 t}{2}}

y = \frac{1}{3} e^{\frac{3 t}{2}} - \frac{1}{3} e^{- \frac{3 t}{2}}

4 x^{^{''}} + 20 x^{^{'}} + 169 x = 0, x (0) = 3, x^{^{'}} (0) = 12

\frac{4}{32} y^{^{''}} + 3 y = 0

y^{^{''}} + 4 y^{^{'}} + 13 y = 0, y^{^{'}} (\frac{π}{2}) = 12

y^{^{''}} + 6 y^{^{'}} + 2 y = 0, y (0) = 4, y^{^{'}} (0) = 0

100 y^{^{''}} - 140 y^{^{'}} + 74 y = 0, y (3) = 6, y^{^{'}} (3) = 5

What is the solution for 36y^{''}-81y=0,y(0)=0,y^'(0)=1 ?
The solution for 36y^{''}-81y=0,y(0)=0,y^'(0)=1 is y= 1/3 e^{(3t)/2}-1/3 e^{-(3t)/2}