What is the solution for y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=0 ?

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

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y^{''}+3y^'+2y=0,y(0)=1,y^'(0)=0

y^{^{''}} + 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = 2 e^{- t} - e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = 2 e^{- t} - e^{- 2 t}

y = 2 e^{- t} - e^{- 2 t}

y = - x^{2} + 7 x, y = x^{2} - 3 x

n = 5 \sum \infty \frac{6 ^{n}}{7 ^{n}}

x \to 0 + lim ((5 x)^{4 x})

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

\int \frac{2}{2 x ^{2} + 4 x - 6} d x

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