What is the solution for 10y^{''}-y^'-3y=0,y(0)=1,y^'(0)=0 ?

The solution for 10y^{''}-y^'-3y=0,y(0)=1,y^'(0)=0 is y=(1-6/11)e^{(3t)/5}+6/11 e^{-t/2}

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10y^{''}-y^'-3y=0,y(0)=1,y^'(0)=0

10 y^{^{''}} - y^{^{'}} - 3 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = (1 - \frac{6}{11}) e^{\frac{3 t}{5}} + \frac{6}{11} e^{- \frac{t}{2}}

Solution steps

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

Solve linear ODE:

y = (1 - \frac{6}{11}) e^{\frac{3 t}{5}} + \frac{6}{11} e^{- \frac{t}{2}}

y = (1 - \frac{6}{11}) e^{\frac{3 t}{5}} + \frac{6}{11} e^{- \frac{t}{2}}

\frac{d}{d x} (\frac{1 - 0.6 x}{x})

\frac{\partial}{\partial x} (\frac{x}{1 + x y})

\frac{d}{d x} (3 x + 6)

y^{^{'}} - 3 y = 6

f (x) = cos (sin (x^{2}))

What is the solution for 10y^{''}-y^'-3y=0,y(0)=1,y^'(0)=0 ?
The solution for 10y^{''}-y^'-3y=0,y(0)=1,y^'(0)=0 is y=(1-6/11)e^{(3t)/5}+6/11 e^{-t/2}