Преобразование Лапласа (операционное исчисление) и его применение к решению дифференциальных уравнений

$f(t) = \sin t \sin 3 t$;

$f(t) = t \operatorname {sh} t ;$

$f(t) = \sin^{3} t$;

$f(t) = \cos^{4} t+\sin^{4} t$.

$F(p) = \frac{1}{p^{2}-4 p+5}$;

$F(p) = \frac{1}{p^{2}\left(p^{2}+1\right)}$;

$F(p) = \frac{p+1}{p^{3}-2 p^{2}-5 p+6}$;

$F(p) = \frac{1}{p(p-1)\left(p^{2}+4\right)}$;

$F(p) = \frac{1}{(p+1)(p+3)^{3}}$;

$F(p) = \frac{p^{2}+2 p-1}{p^{3}+3 p^{2}+3 p+1}$;

$F(p) = \frac{(5 p+3)}{(p-1)\left(p^{2}+2 p+5\right)}$;

$F(p) = \frac{p^{2}}{\left(p^{2}+4\right)\left(p^{2}+9\right)}$;

$F(p) = \frac{1}{(p+1)^{3}(p+3)}$;

$F(p) = \frac{1}{(p-1)^{2}(p-2)^{3}}$.

$x^{\prime \prime }(t)-2 x^{\prime }(t)+x(t) = t^{2} e^{t}, x(0) = 0, x^{\prime }(0) = 1$;

$x^{(3)}(t)+3 x^{(2)}(t)+3 x^{\prime }(t)+x(t) = 1, x(0) = x^{\prime }(0) = x^{\prime \prime }(0) = 0$;

$x^{(3)}(t)-2 x^{(2)}(t)+x^{\prime }(t) = 4, x(0) = 1, x^{\prime }(0) = 2, x^{\prime \prime }(0) = -2$;

$x^{(3)}(t)+x^{(2)}(t) = \cos t, x(0) = -2, x^{\prime }(0) = x^{\prime \prime }(0) = 0$;

$x^{\prime \prime }(t)+x(t) = 2 \cos t, x(0) = -1, x^{\prime }(0) = 1$;

$x^{\prime \prime }(t)+4 x(t) = 2 \sin 2 t, x(0) = -1, x^{\prime }(0) = 0$;

$x^{(4)}(t)-x(t) = 1, x(0) = 1, x^{\prime }(0) = 1, x^{\prime \prime }(0) = 2, x^{(3)}(0) = -1$;

$x^{(4)}(t)-5 x^{(2)}(t)+10 x^{\prime }(t)-6 x(t) = 0$, $x(0) = 1, x^{\prime }(0) = 0, x^{\prime \prime }(0) = 6, x^{(3)}(0) = -14$.

$\begin{cases} x^{\prime }(t) = -x(t)+2 y(t), \\ y^{\prime }(t) = -x(t)-4 y(t),\end{cases} \quad x(0) = y(0) = 1 ;$

$\begin{cases} x^{\prime }(t)-2 x(t)-4 y(t) = 0, \\ y^{\prime }(t)+x(t)+2 y(t) = 0,\end{cases} \quad x(0) = 2, \quad y(0) = 1$;

$\begin{cases} x^{\prime }(t)+3 x(t)-4 y(t) = 9 e^{2 t}, \\ 2 x(t)+y^{\prime }(t)-3 y(t) = 3 e^{2 t},\end{cases} \quad x(0) = 2, \quad y(0) = 0$;

$\begin{cases} x^{\prime }(t)+y^{\prime }(t) = 1, \\ x^{(2)}(t)+y^{\prime }(t)+x(t) = e^{t},\end{cases} x(0) = 1, x^{\prime }(0) = 2, y(0) = 0, y^{\prime }(0) = -1$.