What is the solution for (d^2x)/(dt^2)+4x=t^2 ?

The solution for (d^2x)/(dt^2)+4x=t^2 is x=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(d^2x)/(dt^2)+4x=t^2

\frac{d ^{2} x}{d t ^{2}} + 4 x = t^{2}

x = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

Solution steps

\frac{d ^{2} x}{d t ^{2}} + 4 x = t^{2}

Solve linear ODE:

x = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

x = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{t ^{2}}{4} - \frac{1}{8}

((5 - 120 x)^{2} + (150 x)^{2})

\int_{0}^{1} (x + 1) d x

\int - x cos (x + \frac{π}{3}) + x cos (x) d x

\int \frac{1}{y ( 1 - \frac{y}{m} )} d y

\int 5 t^{4} d t

What is the solution for (d^2x)/(dt^2)+4x=t^2 ?
The solution for (d^2x)/(dt^2)+4x=t^2 is x=c_{1}cos(2t)+c_{2}sin(2t)+(t^2)/4-1/8