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

The solution for y^{''}-6y^'+5y=x,y(0)=e^x is y=c_{1}e^{5x}+c_{2}e^x+x/5+6/25

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

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

y = c_{1} e^{5 x} + c_{2} e^{x} + \frac{x}{5} + \frac{6}{25}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{x} + \frac{x}{5} + \frac{6}{25}

y = c_{1} e^{5 x} + c_{2} e^{x} + \frac{x}{5} + \frac{6}{25}

y^{^{'''}} - 2 y^{^{''}} - y^{^{'}} + 2 y = 6 e^{2 x}

\frac{d ^{2} y}{d x ^{2}} + y = sin (2 x)

y^{^{''}} + 9 y = cos (3 t), y (0) = 2, y^{^{'}} (0) = 5

y^{^{''}} - y^{^{'}} - 6 y = (x^{2} + 3 x) e^{2 x} + cos (2 x)

\frac{d ^{2} y}{d x ^{2}} + \frac{d y}{d x} + 12 y = 20 cos (3 x)

What is the solution for y^{''}-6y^'+5y=x,y(0)=e^x ?
The solution for y^{''}-6y^'+5y=x,y(0)=e^x is y=c_{1}e^{5x}+c_{2}e^x+x/5+6/25