What is the solution for y^{''}-3y^'+2y=11e^{3x},y(0)=e^x ?

The solution for y^{''}-3y^'+2y=11e^{3x},y(0)=e^x is y=c_{1}e^{2x}+c_{2}e^x+11/2 e^{3x}

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y^{''}-3y^'+2y=11e^{3x},y(0)=e^x

y^{^{''}} - 3 y^{^{'}} + 2 y = 11 e^{3 x}, y (0) = e^{x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{11}{2} e^{3 x}

Solution steps

y^{''} (x) - 3 y^{'} (x) + 2 y = 11 e^{3 x}

Solve linear ODE:

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{11}{2} e^{3 x}

y = c_{1} e^{2 x} + c_{2} e^{x} + \frac{11}{2} e^{3 x}

y^{^{''}} + 16 y = e^{4 t} + 9 cos (2 t)

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

\frac{d ^{2} y}{d x ^{2}} - 4 y^{^{'}} - 12 y = 2 x + 6

y^{^{'''}} - 3 y^{^{''}} + 4 y = 4 - e^{- 3 x}

x^{^{''}} + 5 x^{^{'}} + 4 x = 8 e^{t}

What is the solution for y^{''}-3y^'+2y=11e^{3x},y(0)=e^x ?
The solution for y^{''}-3y^'+2y=11e^{3x},y(0)=e^x is y=c_{1}e^{2x}+c_{2}e^x+11/2 e^{3x}