What is the solution for y^{''}-8y^'+16y=e^{4x}ln(9+x^2) ?

The solution for y^{''}-8y^'+16y=e^{4x}ln(9+x^2) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-8y^'+16y=e^{4x}ln(9+x^2)

y^{^{''}} - 8 y^{^{'}} + 16 y = e^{4 x} ln (9 + x^{2})

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{1}{2} e^{4 x} x^{2} ln (x^{2} + 9) - \frac{3}{2} e^{4 x} x^{2} + 6 e^{4 x} x arctan (\frac{x}{3}) - \frac{9}{2} e^{4 x} ln (x^{2} + 9) + \frac{9}{2} e^{4 x}

Solution steps

y^{''} (x) - 8 y^{'} (x) + 16 y = e^{4 x} ln (9 + x^{2})

Solve linear ODE:

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{1}{2} e^{4 x} x^{2} ln (x^{2} + 9) - \frac{3}{2} e^{4 x} x^{2} + 6 e^{4 x} x arctan (\frac{x}{3}) - \frac{9}{2} e^{4 x} ln (x^{2} + 9) + \frac{9}{2} e^{4 x}

y = c_{1} e^{4 x} + c_{2} x e^{4 x} + \frac{1}{2} e^{4 x} x^{2} ln (x^{2} + 9) - \frac{3}{2} e^{4 x} x^{2} + 6 e^{4 x} x arctan (\frac{x}{3}) - \frac{9}{2} e^{4 x} ln (x^{2} + 9) + \frac{9}{2} e^{4 x}

\frac{d ^{2} y}{d x ^{2}} - 8 \frac{d y}{d x} + 16 y = e^{4 x}

y^{^{''}} + y = 5 x sin (x)

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

y^{^{''}} - y^{^{'}} - 2 y = cosh (2 t)

y^{^{''}} - y = sin (x)

What is the solution for y^{''}-8y^'+16y=e^{4x}ln(9+x^2) ?
The solution for y^{''}-8y^'+16y=e^{4x}ln(9+x^2) is