What is the solution for y^{''}-6y^'+25y=50t^3-36t^2-63t+18 ?

The solution for y^{''}-6y^'+25y=50t^3-36t^2-63t+18 is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+2t^3-3t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

y^{''}-6y^'+25y=50t^3-36t^2-63t+18

y^{^{''}} - 6 y^{^{'}} + 25 y = 50 t^{3} - 36 t^{2} - 63 t + 18

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 t^{3} - 3 t

Solution steps

y^{''} (t) - 6 y^{'} (t) + 25 y = 50 t^{3} - 36 t^{2} - 63 t + 18

Solve linear ODE:

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 t^{3} - 3 t

y = e^{3 t} (c_{1} cos (4 t) + c_{2} sin (4 t)) + 2 t^{3} - 3 t

\frac{\partial}{\partial x} (ln (t))

x \to + 2 lim (\frac{x - 7}{x + 3})

\frac{2 x - 7}{5 x + 4} + x^{3}

\int \frac{8}{x ^{2} x ^{2} + 4} d x

\frac{d}{d x} (\frac{e ^{y}}{x + y})

What is the solution for y^{''}-6y^'+25y=50t^3-36t^2-63t+18 ?
The solution for y^{''}-6y^'+25y=50t^3-36t^2-63t+18 is y=e^{3t}(c_{1}cos(4t)+c_{2}sin(4t))+2t^3-3t