What is the solution for 6y^{''}-y^'-y=0,y(0)=10,y^'(0)=3 ?

The solution for 6y^{''}-y^'-y=0,y(0)=10,y^'(0)=3 is y=(10-12/5)e^{t/2}+12/5 e^{-t/3}

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6y^{''}-y^'-y=0,y(0)=10,y^'(0)=3

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

y = (10 - \frac{12}{5}) e^{\frac{t}{2}} + \frac{12}{5} e^{- \frac{t}{3}}

Solution steps

6 y^{''} (t) - y^{'} (t) - y = 0

Solve linear ODE:

y = (10 - \frac{12}{5}) e^{\frac{t}{2}} + \frac{12}{5} e^{- \frac{t}{3}}

y = (10 - \frac{12}{5}) e^{\frac{t}{2}} + \frac{12}{5} e^{- \frac{t}{3}}

\frac{d}{d x} (\frac{x - x}{x + x})

\int \frac{1 + cos ( x )}{2} d x

n = 3 \sum \infty - \frac{5}{n ^{1.01}}

f (x) = 7 x^{3} - 5 x + 8

cos (3 t)

What is the solution for 6y^{''}-y^'-y=0,y(0)=10,y^'(0)=3 ?
The solution for 6y^{''}-y^'-y=0,y(0)=10,y^'(0)=3 is y=(10-12/5)e^{t/2}+12/5 e^{-t/3}