What is the solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=1 ?

The solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=1 is y=-2/27 e^{3t}+(10e^{3t}t)/9+t/9+2/27

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y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=1

y^{^{''}} - 6 y^{^{'}} + 9 y = t, y (0) = 0, y^{^{'}} (0) = 1

y = - \frac{2}{27} e^{3 t} + \frac{10 e ^{3 t} t}{9} + \frac{t}{9} + \frac{2}{27}

Solution steps

y^{''} (t) - 6 y^{'} (t) + 9 y = t

Solve linear ODE:

y = - \frac{2}{27} e^{3 t} + \frac{10 e ^{3 t} t}{9} + \frac{t}{9} + \frac{2}{27}

y = - \frac{2}{27} e^{3 t} + \frac{10 e ^{3 t} t}{9} + \frac{t}{9} + \frac{2}{27}

\frac{d}{d x} (\frac{1}{x} - \frac{1}{sin ( x )})

\frac{d y}{d x} = \frac{6 x ^{4} e ^{\frac{y}{x}} + x ^{2} y ^{2}}{x ^{3} y}

x \to 0 + lim (- x)

\int sec (x) tan (x) - 6 e^{x} d x

y^{^{'}} = 5 (y - t^{2})

What is the solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=1 ?
The solution for y^{''}-6y^'+9y=t,y(0)=0,y^'(0)=1 is y=-2/27 e^{3t}+(10e^{3t}t)/9+t/9+2/27