What is the solution for y^{''}+4y=3t+2 ?

The solution for y^{''}+4y=3t+2 is y=c_{1}cos(2t)+c_{2}sin(2t)+(3t)/4+1/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+4y=3t+2

y^{^{''}} + 4 y = 3 t + 2

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 t}{4} + \frac{1}{2}

Solution steps

y^{''} (t) + 4 y = 3 t + 2

Solve linear ODE:

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 t}{4} + \frac{1}{2}

y = c_{1} cos (2 t) + c_{2} sin (2 t) + \frac{3 t}{4} + \frac{1}{2}

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

y^{^{'''}} + y^{^{''}} = x^{2}

y^{^{'''}} - 2 y^{^{''}} = 3 (y^{^{'}} + x^{2} - 1)

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

y^{^{''}} - y = 2 x^{4} + 5 x^{2} - 10 x

What is the solution for y^{''}+4y=3t+2 ?
The solution for y^{''}+4y=3t+2 is y=c_{1}cos(2t)+c_{2}sin(2t)+(3t)/4+1/2