What is the solution for y^{''}+9y=3t^2 ?

The solution for y^{''}+9y=3t^2 is y=c_{1}cos(3t)+c_{2}sin(3t)+(t^2)/3-2/27

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+9y=3t^2

y^{^{''}} + 9 y = 3 t^{2}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t ^{2}}{3} - \frac{2}{27}

Solution steps

y^{''} (t) + 9 y = 3 t^{2}

Solve linear ODE:

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t ^{2}}{3} - \frac{2}{27}

y = c_{1} cos (3 t) + c_{2} sin (3 t) + \frac{t ^{2}}{3} - \frac{2}{27}

π \cdot x^{\frac{1}{2}} - x^{\frac{1}{3}}

- 4 sin (x)

\frac{d}{d b} (sin^{4} (2 b) - cos^{4} (2 b))

2 x^{2} y - x^{3} y^{^{'}} = y^{3}

\int \frac{x ^{2}}{1 + x ^{6}} d x

What is the solution for y^{''}+9y=3t^2 ?
The solution for y^{''}+9y=3t^2 is y=c_{1}cos(3t)+c_{2}sin(3t)+(t^2)/3-2/27