What is the solution for y^'-y=e^tcos(t),y(0)=1 ?

The solution for y^'-y=e^tcos(t),y(0)=1 is y=e^tsin(t)+e^t

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'-y=e^tcos(t),y(0)=1

y^{^{'}} - y = e^{t} cos (t), y (0) = 1

y = e^{t} sin (t) + e^{t}

Solution steps

y^{'} (t) - y = e^{t} cos (t)

Solve linear ODE:

y = e^{t} sin (t) + e^{t}

y = e^{t} sin (t) + e^{t}

\frac{d}{d x} (\frac{5}{x} - \frac{2}{x ^{3}} + \frac{1}{x ^{4}})

\frac{d}{d x} (sin (2 x) - 2 sin (x))

\int cos (x) cot (x) d x

\frac{d}{d x} (x^{2} + e^{2 x y})

(- 1, 0), (1, 2)

What is the solution for y^'-y=e^tcos(t),y(0)=1 ?
The solution for y^'-y=e^tcos(t),y(0)=1 is y=e^tsin(t)+e^t