Odpowiedź:
[tex]$y(x)=\frac{2}{11} e^{-x}+\frac{139}{22} e^{-\frac{13}{2}x }[/tex]
Szczegółowe wyjaśnienie:
Równanie:
[tex]$y'+\frac{13}{2}y=e^{-x}\\[/tex]
Niech:
[tex]\mathcal{L}[y(x)]=F(t)\\[/tex]
Nakładamy transformatę Laplace'a:
[tex]$\mathcal{L}[y'(x)]+\frac{13}{2} \mathcal{L}[y(x)]=\mathcal{L}[e^{-x}]\\[/tex]
[tex]$t \cdot \mathcal{L}[y(x)] -t^{0} \cdot \frac{13}{2} +\frac{13}{2} \mathcal{L}[y(x)]=\mathcal{L}[e^{-x}]\\[/tex]
[tex]$t \cdot F(t) -\frac{13}{2} +\frac{13}{2} F(t)=\frac{1}{t+1} \\[/tex]
Teraz rozwiązujemy równanie ze względu na [tex]F(t)[/tex] :
[tex]$F(t)(t+\frac{13}{2} )=\frac{1}{t+1}+\frac{13}{2}\\[/tex]
[tex]$F(t)(\frac{2t+13}{2} )=\frac{13t+15}{2t+2} \\[/tex]
[tex]$F(t)=\frac{13t+15}{2t+2} \cdot \frac{2}{2t+13} =\frac{13t+15}{(t+1)(2t+13)}[/tex]
[tex]$F(t)=\frac{\frac{2}{11} }{t+1} +\frac{\frac{139}{11} }{2t+13}[/tex]
Teraz nakładamy transformatę odwrotną:
[tex]$y(x)=\mathcal{L}^{-1}[\frac{\frac{2}{11} }{t+1} +\frac{\frac{139}{11} }{2t+13}][/tex]
[tex]$y(x)=\mathcal{L}^{-1}[\frac{\frac{2}{11} }{t+1}]+\mathcal{L}^{-1}[\frac{\frac{139}{11} }{2t+13}][/tex]
[tex]$y(x)=\frac{2}{11} \mathcal{L}^{-1}[\frac{1}{t+1}]+\frac{139}{11}\mathcal{L}^{-1}[\frac{1 }{2t+13}][/tex]
[tex]$y(x)=\frac{2}{11} \mathcal{L}^{-1}[\frac{1}{t+1}]+\frac{139}{11}\mathcal{L}^{-1}[\frac{\frac{1}{2} }{t+\frac{13}{2} }][/tex]
[tex]$y(x)=\frac{2}{11} \mathcal{L}^{-1}[\frac{1}{t+1}]+\frac{139}{22}\mathcal{L}^{-1}[\frac{1 }{t+\frac{13}{2} }][/tex]
[tex]$y(x)=\frac{2}{11} e^{-x}+\frac{139}{22} e^{-\frac{13}{2}x }[/tex]
