laplacetransform sin(t+c)

解答

拉普拉斯变换

sin (t + c)

\frac{s sin ( c ) + cos ( c )}{s ^{2} + 1}

求解步骤

L {sin (t + c)}

使用拉普拉斯变换表:

L {sin (a t + b)} = \frac{s \cdot sin ( b ) + a \cdot cos ( b )}{s ^{2} + a ^{2}}

L {sin (t + c)} = \frac{s sin ( c ) + 1 \cdot cos ( c )}{s ^{2} + 1 ^{2}}

= \frac{s sin ( c ) + 1 \cdot cos ( c )}{s ^{2} + 1 ^{2}}

= \frac{s sin ( c ) + cos ( c )}{s ^{2} + 1}

\int \frac{x + 3}{x ^{2} 2 x + 3} d x

y^{^{'}} = \frac{x ^{4} y ^{4} - 2 x ^{4}}{y ^{3}}

d er i v a t i v e \frac{x ^{2} - 2 x - 8}{x + 2}

d er i v a t i v e f (x) = \frac{1}{( x ^{3} + 8 ) ^{5}}

\frac{d}{d x} (\frac{sec ( x )}{8 + sec ( x )})