What is the solution for y^{''}-y^'=5sin(2x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^{''}-y^'=5sin(2x)

y^{^{''}} - y^{^{'}} = 5 sin (2 x)

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

Solution steps

y^{''} (x) - y^{'} (x) = 5 sin (2 x)

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (yz e^{x z} - 8)

\int 150 x + 7500 d x

\frac{\partial}{\partial y} (2 x + 3)

\int_{0}^{0.6} 8000 x d x

3 x + 4

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