What is the solution for y^{''}+y-5=e^{2x}-3cos(3x) ?

The solution for y^{''}+y-5=e^{2x}-3cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+1/5 e^{2x}+3/8 cos(3x)+5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}+y-5=e^{2x}-3cos(3x)

y^{^{''}} + y - 5 = e^{2 x} - 3 cos (3 x)

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{5} e^{2 x} + \frac{3}{8} cos (3 x) + 5

Solution steps

y^{''} (x) + y - 5 = e^{2 x} - 3 cos (3 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{5} e^{2 x} + \frac{3}{8} cos (3 x) + 5

y = c_{1} cos (x) + c_{2} sin (x) + \frac{1}{5} e^{2 x} + \frac{3}{8} cos (3 x) + 5

y^{^{'''}} - 18 y^{^{''}} + 864 y = e^{12 x}

(D^{3} - D) y = x

y^{^{''}} + y^{^{'}} = 16 e^{3 x}

\frac{d ^{2} y}{d x ^{2}} + 3 \frac{d y}{d x} + 2 y = \frac{1}{1 + e ^{x}}

y^{^{''}} - y^{^{'}} + 169 y = 13 sin (13 t)

What is the solution for y^{''}+y-5=e^{2x}-3cos(3x) ?
The solution for y^{''}+y-5=e^{2x}-3cos(3x) is y=c_{1}cos(x)+c_{2}sin(x)+1/5 e^{2x}+3/8 cos(3x)+5