What is the solution for y^'-y=8e^x+18e^{4x} ?

The solution for y^'-y=8e^x+18e^{4x} is y=8e^xx+6e^{4x}+c_{1}e^x

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y^'-y=8e^x+18e^{4x}

y^{^{'}} - y = 8 e^{x} + 18 e^{4 x}

y = 8 e^{x} x + 6 e^{4 x} + c_{1} e^{x}

Solution steps

y^{'} (x) - y = 8 e^{x} + 18 e^{4 x}

Solve linear ODE:

y = 8 e^{x} x + 6 e^{4 x} + c_{1} e^{x}

y = 8 e^{x} x + 6 e^{4 x} + c_{1} e^{x}

What is the solution for y^'-y=8e^x+18e^{4x} ?
The solution for y^'-y=8e^x+18e^{4x} is y=8e^xx+6e^{4x}+c_{1}e^x