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

The solution for (dy)/(dt)=sin(2pi(t+3)) is y=-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 >

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

\frac{d y}{d t} = sin (2 π (t + 3))

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

Solution steps

\frac{d y}{d t} = sin (2 π (t + 3))

Substitute

\frac{d y}{d t}

with

y^{'} (t)

y^{^{'}} (t) = sin (2 π (t + 3))

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

then

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

y = \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}

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

\int_{0}^{2} 2 π y (2 y - y^{2}) d y

e^{3 x}

x (\frac{d y}{d x}) = 4 y

{6 e^{- 3 t} - t^{2} + 2 t - 8}

(y - x) e^{x} + (1 + e^{x}) \frac{d y}{d x} = 0

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