What is expand te^{at}(cos(bt)+sin(bt)) ?

The solution to expand te^{at}(cos(bt)+sin(bt)) is e^{ta}tcos(tb)+e^{ta}tsin(tb)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

expand te^{at}(cos(bt)+sin(bt))

expand

t e^{a t} (cos (b t) + sin (b t))

e^{t a} t cos (t b) + e^{t a} t sin (t b)

Solution steps

t e^{a t} (cos (b t) + sin (b t))

= e^{t a} t (cos (t b) + sin (t b))

Apply the distributive law:

a (b + c) = ab + a c

a = t e^{a t}, b = cos (b t), c = sin (b t)

= t e^{a t} cos (b t) + t e^{a t} sin (b t)

= e^{t a} t cos (t b) + e^{t a} t sin (t b)

\frac{( - x + 14 ) \cdot ( x - 6 )}{2}

\frac{( 3 x )}{( x + 8 ) ^{2}}

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

(d - a)^{2} + (c)^{2}

(\frac{( 2 + 2 i )}{2})^{2}

What is expand te^{at}(cos(bt)+sin(bt)) ?
The solution to expand te^{at}(cos(bt)+sin(bt)) is e^{ta}tcos(tb)+e^{ta}tsin(tb)