What is the solution for y^{''}+3y^'+0=5 ?

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

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y^{''}+3y^'+0=5

y^{^{''}} + 3 y^{^{'}} + 0 = 5

y = c_{1} + c_{2} e^{- 3 t} + \frac{5 t}{3}

Solution steps

y^{''} (t) + 3 y^{'} (t) + 0 = 5

Solve linear ODE:

y = c_{1} + c_{2} e^{- 3 t} + \frac{5 t}{3}

y = c_{1} + c_{2} e^{- 3 t} + \frac{5 t}{3}

\int_{0}^{\frac{1}{3}} \frac{2}{1 + 9 x ^{2}} d x

f (x) = x^{3} - 16 x + 5, at x = 4

\int_{0}^{2} \frac{1}{x ^{2} - 5 x + 4} d x

\frac{d}{d x} (\frac{2}{1 + e ^{- x}})

n = 0 \sum \infty 3 (- \frac{1}{2})^{n}

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