What is the solution for y^'+4y=cos(x) ?

The solution for y^'+4y=cos(x) is y=(sin(x))/(17)+(4cos(x))/(17)+(c_{1})/(e^{4x)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+4y=cos(x)

y^{^{'}} + 4 y = cos (x)

y = \frac{sin ( x )}{17} + \frac{4 cos ( x )}{17} + \frac{c _{1}}{e ^{4 x}}

Solution steps

y^{'} (x) + 4 y = cos (x)

Solve linear ODE:

y = \frac{sin ( x )}{17} + \frac{4 cos ( x )}{17} + \frac{c _{1}}{e ^{4 x}}

y = \frac{sin ( x )}{17} + \frac{4 cos ( x )}{17} + \frac{c _{1}}{e ^{4 x}}

ln (\frac{x}{2}), at 2

\frac{d y}{d x} = \frac{5 y}{2 x}

4 x^{5}, at (a = 3)

6 x - 18, [2, 6]

What is the solution for y^'+4y=cos(x) ?
The solution for y^'+4y=cos(x) is y=(sin(x))/(17)+(4cos(x))/(17)+(c_{1})/(e^{4x)}