What is the solution for 2y^{''}-7y^'+5y=0,y(0)=4,y^'(0)=5.5 ?

The solution for 2y^{''}-7y^'+5y=0,y(0)=4,y^'(0)=5.5 is y=e^{(5t)/2}+3e^t

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2y^{''}-7y^'+5y=0,y(0)=4,y^'(0)=5.5

2 y^{^{''}} - 7 y^{^{'}} + 5 y = 0, y (0) = 4, y^{^{'}} (0) = 5.5

y = e^{\frac{5 t}{2}} + 3 e^{t}

Solution steps

2 y^{''} (t) - 7 y^{'} (t) + 5 y = 0

Solve linear ODE:

y = e^{\frac{5 t}{2}} + 3 e^{t}

y = e^{\frac{5 t}{2}} + 3 e^{t}

y^{^{''}} - 6 y^{^{'}} - 7 y = 0, y (0) = 5, y^{^{'}} (0) = - 5

\frac{d ^{2} y}{d t ^{2}} + 18 \frac{d y}{d t} + 81 y = 0

6 y^{^{''}} + 120 y^{^{'}} + 50 y = 0

\frac{d ^{2} q}{d t ^{2}} + q^{^{'}} + q = 0

16 \cdot y^{^{'''}} - y^{^{'}} = 0

What is the solution for 2y^{''}-7y^'+5y=0,y(0)=4,y^'(0)=5.5 ?
The solution for 2y^{''}-7y^'+5y=0,y(0)=4,y^'(0)=5.5 is y=e^{(5t)/2}+3e^t