What is the solution for y^{''}-8y^'+7y=4e^3x ?

The solution for y^{''}-8y^'+7y=4e^3x is y=c_{1}e^{7x}+c_{2}e^x+(4e^3x)/7+(32e^3)/(49)

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y^{''}-8y^'+7y=4e^3x

y^{^{''}} - 8 y^{^{'}} + 7 y = 4 e^{3} x

y = c_{1} e^{7 x} + c_{2} e^{x} + \frac{4 e ^{3} x}{7} + \frac{32 e ^{3}}{49}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 7 y = 4 e^{3} x

Solve linear ODE:

y = c_{1} e^{7 x} + c_{2} e^{x} + \frac{4 e ^{3} x}{7} + \frac{32 e ^{3}}{49}

y = c_{1} e^{7 x} + c_{2} e^{x} + \frac{4 e ^{3} x}{7} + \frac{32 e ^{3}}{49}

What is the solution for y^{''}-8y^'+7y=4e^3x ?
The solution for y^{''}-8y^'+7y=4e^3x is y=c_{1}e^{7x}+c_{2}e^x+(4e^3x)/7+(32e^3)/(49)