What is the solution for 2y^{''}+3y^'-3y=0 ?

The solution for 2y^{''}+3y^'-3y=0 is y=c_{1}e^{((-3+sqrt(33))t)/4}+c_{2}e^{((-3-sqrt(33))t)/4}

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2y^{''}+3y^'-3y=0

2 y^{^{''}} + 3 y^{^{'}} - 3 y = 0

y = c_{1} e^{\frac{( - 3 + 33 ) t}{4}} + c_{2} e^{\frac{( - 3 - 33 ) t}{4}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 3 + 33 ) t}{4}} + c_{2} e^{\frac{( - 3 - 33 ) t}{4}}

y = c_{1} e^{\frac{( - 3 + 33 ) t}{4}} + c_{2} e^{\frac{( - 3 - 33 ) t}{4}}

\frac{d ^{2} y}{d x ^{2}} + 4 y = 0, y = 0

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

y^{^{''}} - \frac{3}{2} y^{^{'}} + y = 0

77 y^{^{''''}} + 39 y^{^{''}} + 4 y = 0

\frac{d ^{2} y}{d t ^{2}} + 2 \frac{d y}{d t} + 10 y = 0

What is the solution for 2y^{''}+3y^'-3y=0 ?
The solution for 2y^{''}+3y^'-3y=0 is y=c_{1}e^{((-3+sqrt(33))t)/4}+c_{2}e^{((-3-sqrt(33))t)/4}