What is the solution for y^{''}-y^'-12y=-5,y(0)=0,y^'(0)=0 ?

The solution for y^{''}-y^'-12y=-5,y(0)=0,y^'(0)=0 is y=(5/21-5/12)e^{4t}-5/21 e^{-3t}+5/12

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y^{''}-y^'-12y=-5,y(0)=0,y^'(0)=0

y^{^{''}} - y^{^{'}} - 12 y = - 5, y (0) = 0, y^{^{'}} (0) = 0

y = (\frac{5}{21} - \frac{5}{12}) e^{4 t} - \frac{5}{21} e^{- 3 t} + \frac{5}{12}

Solution steps

y^{''} (t) - y^{'} (t) - 12 y = - 5

Solve linear ODE:

y = (\frac{5}{21} - \frac{5}{12}) e^{4 t} - \frac{5}{21} e^{- 3 t} + \frac{5}{12}

y = (\frac{5}{21} - \frac{5}{12}) e^{4 t} - \frac{5}{21} e^{- 3 t} + \frac{5}{12}

x \to 2 - lim (\frac{x}{2 - x})

\int \frac{x ^{2}}{289 - x ^{2}} d x

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

(0.4) (2)

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

What is the solution for y^{''}-y^'-12y=-5,y(0)=0,y^'(0)=0 ?
The solution for y^{''}-y^'-12y=-5,y(0)=0,y^'(0)=0 is y=(5/21-5/12)e^{4t}-5/21 e^{-3t}+5/12