What is the solution for y^{''}-9y=e^{3x},y(0)=1,y^'(0)=-5 ?

The solution for y^{''}-9y=e^{3x},y(0)=1,y^'(0)=-5 is y=(1-49/36)e^{3x}+49/36 e^{-3x}+(e^{3x}x)/6

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y^{''}-9y=e^{3x},y(0)=1,y^'(0)=-5

y^{^{''}} - 9 y = e^{3 x}, y (0) = 1, y^{^{'}} (0) = - 5

y = (1 - \frac{49}{36}) e^{3 x} + \frac{49}{36} e^{- 3 x} + \frac{e ^{3 x} x}{6}

Solution steps

y^{''} (x) - 9 y = e^{3 x}

Solve linear ODE:

y = (1 - \frac{49}{36}) e^{3 x} + \frac{49}{36} e^{- 3 x} + \frac{e ^{3 x} x}{6}

y = (1 - \frac{49}{36}) e^{3 x} + \frac{49}{36} e^{- 3 x} + \frac{e ^{3 x} x}{6}

What is the solution for y^{''}-9y=e^{3x},y(0)=1,y^'(0)=-5 ?
The solution for y^{''}-9y=e^{3x},y(0)=1,y^'(0)=-5 is y=(1-49/36)e^{3x}+49/36 e^{-3x}+(e^{3x}x)/6