laplacetransform e^{3t}(9-4t+10sin(t/2))

Lösung

laplace transformation

e^{3 t} (9 - 4 t + 10 sin (\frac{t}{2}))

\frac{9}{s - 3} - \frac{4}{( s - 3 ) ^{2}} + \frac{20}{4 ( s - 3 ) ^{2} + 1}

Schritte zur Lösung

L {e^{3 t} (- 4 t + 10 sin (\frac{t}{2}) + 9)}

Wende Transformationsregel an: wenn

L {f (t)} = F (s)

dann

L {e^{a t} f (t)} = F (s - a)

Für

e^{3 t} (9 - 4 t + 10 sin (\frac{t}{2})) : f (t) = (9 - 4 t + 10 sin (\frac{t}{2})), a = 3

= L {(9 - 4 t + 10 sin (\frac{t}{2}))} (s - 3)

L {(9 - 4 t + 10 sin (\frac{t}{2}))} : \frac{9}{s} - \frac{4}{s ^{2}} + \frac{20}{4 s ^{2} + 1}

= \frac{9}{( s - 3 )} - \frac{4}{( s - 3 ) ^{2}} + \frac{20}{4 ( s - 3 ) ^{2} + 1}

= \frac{9}{s - 3} - \frac{4}{( s - 3 ) ^{2}} + \frac{20}{4 ( s - 3 ) ^{2} + 1}

x \to 0 + lim (x ln (e^{x} - 1))

d er i v a t i v e f (x) = \frac{x ^{2} - 99}{x - 10}

f (x) = e^{5 x^{2}}

\frac{d y}{d x} = 2 y + x^{2} + 7

\int_{0}^{\frac{π}{4}} sin^{2} (x) d x