What is the solution for 2y^{''}+7y^'+3y=6,y(0)=0,y^'(0)=0 ?

The solution for 2y^{''}+7y^'+3y=6,y(0)=0,y^'(0)=0 is y=(-2/5-2)e^{-t/2}+2/5 e^{-3t}+2

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2y^{''}+7y^'+3y=6,y(0)=0,y^'(0)=0

2 y^{^{''}} + 7 y^{^{'}} + 3 y = 6, y (0) = 0, y^{^{'}} (0) = 0

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

Solution steps

2 y^{''} (t) + 7 y^{'} (t) + 3 y = 6

Solve linear ODE:

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

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

f (x) = \frac{2}{x + 7}

\int x cos (3) x^{2} d x

\frac{d}{d x} (5 x^{4} + 3 x^{2} - 9 x + 16)

f (x) = 7 e^{x} + x, (0, 7)

\int \frac{2 - v}{v ^{2} - 1} d v

What is the solution for 2y^{''}+7y^'+3y=6,y(0)=0,y^'(0)=0 ?
The solution for 2y^{''}+7y^'+3y=6,y(0)=0,y^'(0)=0 is y=(-2/5-2)e^{-t/2}+2/5 e^{-3t}+2