What is the solution for y^{''}-3y^'=-18x,y(0)=0,y^'(0)=5 ?

The solution for y^{''}-3y^'=-18x,y(0)=0,y^'(0)=5 is y=e^{3x}-1+3x^2+2x

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y^{''}-3y^'=-18x,y(0)=0,y^'(0)=5

y^{^{''}} - 3 y^{^{'}} = - 18 x, y (0) = 0, y^{^{'}} (0) = 5

y = e^{3 x} - 1 + 3 x^{2} + 2 x

Solution steps

y^{''} (x) - 3 y^{'} (x) = - 18 x

Solve linear ODE:

y = e^{3 x} - 1 + 3 x^{2} + 2 x

y = e^{3 x} - 1 + 3 x^{2} + 2 x

y^{^{''}} - 8 y^{^{'}} + 17 y = x^{2} e^{4 x}

y^{^{''}} + 25 y^{^{'}} = 6 sin (x)

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

y^{^{''}} - y^{^{'}} = t^{2}

y^{^{''}} - 3 y^{^{'}} + 2 y = (x^{2} + x) e^{3 x}

What is the solution for y^{''}-3y^'=-18x,y(0)=0,y^'(0)=5 ?
The solution for y^{''}-3y^'=-18x,y(0)=0,y^'(0)=5 is y=e^{3x}-1+3x^2+2x