What is the solution for 5y^{''}-7y^'+2y=0,y(0)=1,y^'(0)=-0.8 ?

The solution for 5y^{''}-7y^'+2y=0,y(0)=1,y^'(0)=-0.8 is y=-2e^t+3e^{(2t)/5}

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5y^{''}-7y^'+2y=0,y(0)=1,y^'(0)=-0.8

5 y^{^{''}} - 7 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = - 0.8

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

Solution steps

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

Solve linear ODE:

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

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

4 x^{^{''}} + x^{^{'}} + 2 x = 0, x (0) = 0, x (0) = - 1, x^{^{'}} (0) = 12

2 y^{^{''}} - 5 y^{^{'}} - 3 y = 0, y (0) = 0, y^{^{'}} (0) = 2

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

35 y^{^{''''}} + 143 y^{^{''}} + 144 y = 0

y^{^{'''}} - 7 y^{^{''}} + 10 y = 0, y^{^{'}} (0) = 1

What is the solution for 5y^{''}-7y^'+2y=0,y(0)=1,y^'(0)=-0.8 ?
The solution for 5y^{''}-7y^'+2y=0,y(0)=1,y^'(0)=-0.8 is y=-2e^t+3e^{(2t)/5}