What is the solution for (dx)/(dt)=sin(2pi(t+3)) ?

The solution for (dx)/(dt)=sin(2pi(t+3)) is x=-1/(2pi)cos(2pi(t+3))+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(dx)/(dt)=sin(2pi(t+3))

\frac{d x}{d t} = sin (2 π (t + 3))

x = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

Solution steps

\frac{d x}{d t} = sin (2 π (t + 3))

Substitute

\frac{d x}{d t}

with

x^{'} (t)

x^{^{'}} (t) = sin (2 π (t + 3))

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

x = \int sin (2 π (t + 3)) d t

\int sin (2 π (t + 3)) d t = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

x = - \frac{1}{2 π} cos (2 π (t + 3)) + c_{1}

1

n = 3 \sum \infty \frac{1}{3} (\frac{2}{5})^{n}

\int \frac{1}{5 ^{4 x}} d x

\frac{d}{d x} (\frac{3 cos ( x )}{3 x ^{2} + 3 x - 1})

What is the solution for (dx)/(dt)=sin(2pi(t+3)) ?
The solution for (dx)/(dt)=sin(2pi(t+3)) is x=-1/(2pi)cos(2pi(t+3))+c_{1}