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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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}

\int 4 sec^{2} (x) tan (x) d x

f (x) = x^{3} - 2

\frac{d}{d x} (lo g_{4} (x^{2} - 7 x))

\int x^{2} + 9 d 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