What is the solution for y^{''}-5y^'+6y=4,e^{4x}y(0)=0,y^'(0)=1 ?

The solution for y^{''}-5y^'+6y=4,e^{4x}y(0)=0,y^'(0)=1 is y=c_{1}e^{3x}+c_{2}e^{2x}+2/3

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y^{''}-5y^'+6y=4,e^{4x}y(0)=0,y^'(0)=1

y^{^{''}} - 5 y^{^{'}} + 6 y = 4, e^{4 x} y (0) = 0, y^{^{'}} (0) = 1

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

Solution steps

y^{''} (x) - 5 y^{'} (x) + 6 y = 4

Solve linear ODE:

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

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

\frac{d ^{2} y}{d x ^{2}} - 5 \frac{d y}{d x} - 24 y = 5 + 2 x

y^{^{''}} + y^{^{'}} - 2 y = x e^{- 2 x} + x e^{x}

\frac{1}{4} y^{^{''}} + 4 y = 2 tan (4 t) - \frac{1}{3} e^{2 t}

y^{^{''}} + 5 y^{^{'}} + 4 y = 1 + x + x^{2}

y^{^{''}} + y = - x^{2}, y (0) = 0, y (1) = 1

What is the solution for y^{''}-5y^'+6y=4,e^{4x}y(0)=0,y^'(0)=1 ?
The solution for y^{''}-5y^'+6y=4,e^{4x}y(0)=0,y^'(0)=1 is y=c_{1}e^{3x}+c_{2}e^{2x}+2/3