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

The solution for y^'+y=sin(t) is y=-(cos(t))/2+(sin(t))/2+(c_{1})/(e^t)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'+y=sin(t)

y^{^{'}} + y = sin (t)

y = - \frac{cos ( t )}{2} + \frac{sin ( t )}{2} + \frac{c _{1}}{e ^{t}}

Solution steps

y^{'} (t) + y = sin (t)

Solve linear ODE:

y = - \frac{cos ( t )}{2} + \frac{sin ( t )}{2} + \frac{c _{1}}{e ^{t}}

y = - \frac{cos ( t )}{2} + \frac{sin ( t )}{2} + \frac{c _{1}}{e ^{t}}

\frac{d}{d x} (3 + ln (x))

\frac{e ^{- 7 s}}{s ^{2}}

\int 2 sin (x^{2}) d x

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

\int 13 tan^{5} (x) sec^{4} (x) d x

What is the solution for y^'+y=sin(t) ?
The solution for y^'+y=sin(t) is y=-(cos(t))/2+(sin(t))/2+(c_{1})/(e^t)