What is the solution for y^{''}+a*y^'=b,y(0)=0,y^'(0)=0 ?

The solution for y^{''}+a*y^'=b,y(0)=0,y^'(0)=0 is y=-b/(a^2)+(be^{-at}}{a^2}+\frac{bt)/a

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y^{''}+a*y^'=b,y(0)=0,y^'(0)=0

y^{^{''}} + a \cdot y^{^{'}} = b, y (0) = 0, y^{^{'}} (0) = 0

y = - \frac{b}{a ^{2}} + \frac{b e ^{- a t}}{a ^{2}} + \frac{b t}{a}

Solution steps

y^{''} (t) + a \cdot y^{'} (t) = b

Solve linear ODE:

y = - \frac{b}{a ^{2}} + \frac{b e ^{- a t}}{a ^{2}} + \frac{b t}{a}

y = - \frac{b}{a ^{2}} + \frac{b e ^{- a t}}{a ^{2}} + \frac{b t}{a}

\frac{d y}{d x} = \frac{( 2 x + sec ^{2} ( x ) )}{2 y}, y (0) = - 6

\int \frac{3 x + 5}{x ^{2} + 1} d x

\int \frac{3 x ^{2} - 6 x - 4}{x ^{3} - 3 x ^{2} - 4 x + 12} d x

\frac{d}{d x} (x sinh (x))

x \to 2 + lim (\frac{x ^{3} - 8}{x - 2})

What is the solution for y^{''}+a*y^'=b,y(0)=0,y^'(0)=0 ?
The solution for y^{''}+a*y^'=b,y(0)=0,y^'(0)=0 is y=-b/(a^2)+(be^{-at}}{a^2}+\frac{bt)/a