What is the solution for y^'+4y=cos(4t) ?

The solution for y^'+4y=cos(4t) is y=(sin(4t))/8+(cos(4t))/8+(c_{1})/(e^{4t)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+4y=cos(4t)

y^{^{'}} + 4 y = cos (4 t)

y = \frac{sin ( 4 t )}{8} + \frac{cos ( 4 t )}{8} + \frac{c _{1}}{e ^{4 t}}

Solution steps

y^{'} (t) + 4 y = cos (4 t)

Solve linear ODE:

y = \frac{sin ( 4 t )}{8} + \frac{cos ( 4 t )}{8} + \frac{c _{1}}{e ^{4 t}}

y = \frac{sin ( 4 t )}{8} + \frac{cos ( 4 t )}{8} + \frac{c _{1}}{e ^{4 t}}

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

\frac{1}{1 - x ^{2}}

\int csc (\frac{v - π}{2}) cot (\frac{v - π}{2}) d v

\frac{d y}{d x} + (\frac{4}{x}) y = - 4 x - 5

\int_{0}^{1} \frac{1}{x ( x + 1 )} d x

What is the solution for y^'+4y=cos(4t) ?
The solution for y^'+4y=cos(4t) is y=(sin(4t))/8+(cos(4t))/8+(c_{1})/(e^{4t)}