What is the solution for y^{''}+4y^'+3=0 ?

The solution for y^{''}+4y^'+3=0 is y=c_{1}+c_{2}e^{-4t}-(3t)/4

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y^{''}+4y^'+3=0

y^{^{''}} + 4 y^{^{'}} + 3 = 0

y = c_{1} + c_{2} e^{- 4 t} - \frac{3 t}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 = 0

Solve linear ODE:

y = c_{1} + c_{2} e^{- 4 t} - \frac{3 t}{4}

y = c_{1} + c_{2} e^{- 4 t} - \frac{3 t}{4}

\frac{d y}{d x} = (x + y - 3)^{2}

cos (x), 0

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

\int x^{3} x^{2} + 36 d x

x \to 0 lim (\frac{( 2 x ^{3} )}{2 x})

What is the solution for y^{''}+4y^'+3=0 ?
The solution for y^{''}+4y^'+3=0 is y=c_{1}+c_{2}e^{-4t}-(3t)/4