What is the solution for y^{''}-8y^'+7y=x ?

The solution for y^{''}-8y^'+7y=x is y=c_{1}e^{7x}+c_{2}e^x+x/7+8/49

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y^{''}-8y^'+7y=x

y^{^{''}} - 8 y^{^{'}} + 7 y = x

y = c_{1} e^{7 x} + c_{2} e^{x} + \frac{x}{7} + \frac{8}{49}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 7 y = x

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} e^{x} + \frac{x}{7} + \frac{8}{49}

y = c_{1} e^{7 x} + c_{2} e^{x} + \frac{x}{7} + \frac{8}{49}

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

2 y^{^{''}} + 4 y^{^{'}} + 7 y = x^{2}

y^{^{''}} + 4 y^{^{'}} = 16 x

y^{^{'''}} - y^{^{''}} - 25 y^{^{'}} + 25 y = 3 - e^{x} + e^{5 x}

y^{^{''}} - 4 y^{^{'}} + 3 y = 2 x + 1

What is the solution for y^{''}-8y^'+7y=x ?
The solution for y^{''}-8y^'+7y=x is y=c_{1}e^{7x}+c_{2}e^x+x/7+8/49