Zad.1

${\ddot{y}}_{(t)}$+$5{\dot{y}}_{(t)}$+6y_(t)=0 y(0⁺)=a $\dot{y}(0^{+})$=b

L{${\ddot{y}}_{(t)}$+$5{\dot{y}}_{(t)}$+6y_(t)}=0

L{${\ddot{y}}_{(t)}$}+5L{$5{\dot{y}}_{(t)}$}+6L{y_(t)}=0 Stosuje 3 twierdzenie Laplasa o transformacie pochodnej:

$s^{2}\left( Y_{\left( s \right)} - \frac{a}{s} - \frac{b}{s^{2}} \right) + 5s\left( Y_{\left( s \right)} - \frac{a}{s} \right) + 6Y_{(s)} = 0$

s²Y_(s) − as − b + 5sY_(s) − as + 6Y_(s) = 0

Y_(s)(s²+5s+6) = 2as + b /: (s²+5s+6)

$Y_{\left( s \right)} = \frac{2as + b\ }{\left( s^{2} + 5s + 6 \right)}$ Liczymy deltę i rozkładamy na ułamki proste:

=25 − 24 = 1

$\sqrt{} = 1$

s₁ = −3

s₂ = −2

$Y_{\left( s \right)} = \frac{2as + b\ }{\left( s + 3 \right)(s + 2)} = \frac{A}{s + 3} + \frac{B}{s + 2}$ /* (s+3)(s + 2)

2as + b = As + 2A + Bs + 3B

s¹ : 2a = A + B

s⁰ : b = 2A + 2B

B = 2a − A

2A + 2a − A = b

A = b − 2a

B = 2a − (b − 2a)

B = 4a + b

$Y_{\left( s \right)} = \frac{b - 2a}{s + 3} + \frac{4a + b}{s + 2}$

$L^{- 1}\left\{ Y_{\left( s \right)} \right\} = L^{- 1}\left\{ \frac{b - 2a}{s + 3} \right\} + L^{- 1}\left\{ \frac{4a + b}{s + 2}\ \right\}$

Y_(t) = ( b−2a)e^−3t + (4a + b)e^−2t dla t ≥ 0

Zad 2.

${\ddot{y}}_{(t)}$+4${\dot{y}}_{(t)}$+8y_(t)=E*1_(t) y(0⁺)=0 $\dot{y}(0^{+})$=0

L{${\ddot{y}}_{(t)}$+4${\dot{y}}_{(t)}$+8y_(t)}=E*L{1_(t)}

L{${\ddot{y}}_{(t)}$}+4L{${\dot{y}}_{(t)}$}+8L{y_(t)}=E$*\frac{1}{s}$ Stosuje 3 twierdzenie Laplasa o transformacie pochodnej:

s²Y_(s)+4sY_(s)+8Y_(s)= E$*\frac{1}{s}$

Y_(s)(s²+4s+8)= E$*\frac{1}{s}$ /: (s²+4s+8)

Y_(s)=$\frac{E}{(s^{2} + 4s + 8)}*\frac{1}{s}$

=b² − 4ac = 4² − 32 =   − 16 Delta jest ujemna więc:

Y_(s)=$\frac{E}{(s^{2} + 4s + 8)}*\frac{1}{s}$ = $\frac{\text{As} + B}{s^{2} + 4s + 8} + \frac{C}{s}$ / *(s²+4s+8)*s

E=(As+B)s+C(s²+4s+8)=As²+Bs+Cs²+4Cs+8C

s²: 0=A+C

s¹: 0=B+4C

s⁰: E=8C /: *

C=$\frac{E}{8}$

A= - C=$\ - \frac{E}{8}$

B= - 4C= $- \frac{E}{2}$ Podstawiamy pod A,B i C:

Y_(s)= $\frac{- \frac{E}{8}s - \frac{E}{2}}{s^{2} + 4s + 8} + \frac{\frac{E}{8}}{s}$

$\frac{- \ \frac{E}{8}s\ - \ \frac{E}{2}}{s^{2} + 4s + 8}$=$\frac{s + 2}{{(s + 2)}^{2} + 2^{2}}*( - \frac{E}{8})$+$\ \frac{2}{{(s + 2)}^{2} + 2^{2}}*( - \frac{E}{8})$

Y_(s)=$\frac{s + 2}{{(s + 2)}^{2} + 2^{2}}*\left( - \frac{E}{8} \right) + \frac{2}{{(s + 2)}^{2} + 2^{2}}*( - \frac{E}{8}) + \frac{\frac{E}{8}}{s}$

L⁻¹{Y_(s)}=L⁻¹{$\frac{s + 2}{{(s + 2)}^{2} + 2^{2}}$}$*\left( - \frac{E}{8} \right)$+ L⁻¹{$\frac{2}{{(s + 2)}^{2} + 2^{2}}$}$*\left( - \frac{E}{8} \right)$ + L⁻¹{$\frac{\frac{E}{8}}{s}$}

Y_(t)=$- \frac{E}{8}\cos{2t{*e}^{- 2t} - \frac{E}{8}}\sin{2te^{- 2t}} + \frac{E}{8}*1_{\left( t \right)}$ dla t≥0