What is the solution for y^{''}-7y^'+10y=e^{3x} ?

The solution for y^{''}-7y^'+10y=e^{3x} is y=c_{1}e^{5x}+c_{2}e^{2x}-1/2 e^{3x}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-7y^'+10y=e^{3x}

y^{^{''}} - 7 y^{^{'}} + 10 y = e^{3 x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} - \frac{1}{2} e^{3 x}

Solution steps

y^{''} (x) - 7 y^{'} (x) + 10 y = e^{3 x}

Solve linear ODE:

y = c_{1} e^{5 x} + c_{2} e^{2 x} - \frac{1}{2} e^{3 x}

y = c_{1} e^{5 x} + c_{2} e^{2 x} - \frac{1}{2} e^{3 x}

\frac{\partial}{\partial x} (2 \cdot z \cdot cos (x))

x \to 0 lim (\frac{2 ln ( x + 1 )}{x})

3 e^{- x^{2}}

y = x^{3} sin^{2} (x)

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

What is the solution for y^{''}-7y^'+10y=e^{3x} ?
The solution for y^{''}-7y^'+10y=e^{3x} is y=c_{1}e^{5x}+c_{2}e^{2x}-1/2 e^{3x}