What is the solution for 2y^{''}-7y^'+3y=0,y(0)=1,y^'(0)=5.5 ?

The solution for 2y^{''}-7y^'+3y=0,y(0)=1,y^'(0)=5.5 is y=2e^{3t}-e^{t/2}

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2y^{''}-7y^'+3y=0,y(0)=1,y^'(0)=5.5

2 y^{^{''}} - 7 y^{^{'}} + 3 y = 0, y (0) = 1, y^{^{'}} (0) = 5.5

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

Solution steps

2 y^{''} (t) - 7 y^{'} (t) + 3 y = 0

Solve linear ODE:

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

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

5 x^{^{''}} + 4.9 x^{^{'}} + 10 x = 0

y^{^{''}} + 0 y^{^{'}} + 64 y = 0, y (0) = 5, y^{^{'}} (0) = 0

121 y^{^{''}} - 88 y^{^{'}} + 41 y = 0, y (0) = 5, y^{^{'}} (0) = 3

21 y^{^{''''}} + 82 y^{^{''}} + 25 y = 0

y^{^{''}} + 6 y^{^{'}} + 10 y = 0, y (0) = 0, y^{^{'}} (0) = 6

What is the solution for 2y^{''}-7y^'+3y=0,y(0)=1,y^'(0)=5.5 ?
The solution for 2y^{''}-7y^'+3y=0,y(0)=1,y^'(0)=5.5 is y=2e^{3t}-e^{t/2}