What is the solution for 2y^{''}-5y^'-3y=0,y(0)=0,y^'(0)=2 ?

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

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

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

y = \frac{4}{7} e^{3 t} - \frac{4}{7} e^{- \frac{t}{2}}

Solution steps

2 y^{''} (t) - 5 y^{'} (t) - 3 y = 0

Solve linear ODE:

y = \frac{4}{7} e^{3 t} - \frac{4}{7} e^{- \frac{t}{2}}

y = \frac{4}{7} e^{3 t} - \frac{4}{7} e^{- \frac{t}{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

7 x^{^{''}} + 3 x^{^{'}} + 7 x = 0, x (0) = - 4, x^{^{'}} (0) = 0

y^{^{'''}} - 21 y^{^{''}} + 128 y^{^{'}} - 240 y = 0

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