What is the solution for y^{''}-6y^'+9y=e^{3x}log_{e}(1+x^2) ?

The solution for y^{''}-6y^'+9y=e^{3x}log_{e}(1+x^2) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-6y^'+9y=e^{3x}log_{e}(1+x^2)

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} lo g_{e} (1 + x^{2})

y = c_{1} e^{3 x} + c_{2} x e^{3 x} + \frac{1}{2} e^{3 x} x^{2} ln (x^{2} + 1) - \frac{3}{2} e^{3 x} x^{2} + 2 e^{3 x} x arctan (x) - \frac{1}{2} e^{3 x} ln (x^{2} + 1) + \frac{1}{2} e^{3 x}

Solution steps

y^{''} (x) - 6 y^{'} (x) + 9 y = e^{3 x} lo g_{e} (1 + x^{2})

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} x e^{3 x} + \frac{1}{2} e^{3 x} x^{2} ln (x^{2} + 1) - \frac{3}{2} e^{3 x} x^{2} + 2 e^{3 x} x arctan (x) - \frac{1}{2} e^{3 x} ln (x^{2} + 1) + \frac{1}{2} e^{3 x}

y = c_{1} e^{3 x} + c_{2} x e^{3 x} + \frac{1}{2} e^{3 x} x^{2} ln (x^{2} + 1) - \frac{3}{2} e^{3 x} x^{2} + 2 e^{3 x} x arctan (x) - \frac{1}{2} e^{3 x} ln (x^{2} + 1) + \frac{1}{2} e^{3 x}

y^{^{''}} - y^{^{'}} = e^{2 x} - 4 x

y^{^{''}} + 2 y^{^{'}} + y = e^{- x} 1 + x

y^{^{''}} - 6 y^{^{'}} + 8 y = 64 x^{2}

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

y^{^{''}} - 2 y^{^{'}} = 4 e^{2 t}

What is the solution for y^{''}-6y^'+9y=e^{3x}log_{e}(1+x^2) ?
The solution for y^{''}-6y^'+9y=e^{3x}log_{e}(1+x^2) is