Laplace Transform - Partial Fractions

Question

Please see attached image. I keep coming up with a irrational/complex coeffecient which is correct. Can you please help me put it partial fractions please

Thanks

Answer 1

• $$x'(t)+5x(t)+2y(t)=e^{-t}\Longleftrightarrow$$ $$\mathcal{L}_t\left[x'(t)+5x(t)+2y(t)\right]_{(s)}=\mathcal{L}_t\left[e^{-t}\right]_{(s)}\Longleftrightarrow$$ $$sx(s)-x(0)+5x(s)+2y(s)=\frac{1}{1+s}\Longleftrightarrow$$ $$x(s)\left[s+5\right]=\frac{1}{1+s}+x(0)-2y(s)\Longleftrightarrow$$ $$x(s)=\frac{\frac{1}{1+s}+x(0)-2y(s)}{s+5}$$

• $$y'(t)+2x(t)+2y(t)=0\Longleftrightarrow$$ $$\mathcal{L}_t\left[y'(t)+2x(t)+2y(t)\right]_{(s)}=\mathcal{L}_t\left[0\right]_{(s)}\Longleftrightarrow$$ $$sy(s)-y(0)+2x(s)+2y(s)=0\Longleftrightarrow$$ $$y(s)\left[s+2\right]=y(0)-2x(s)\Longleftrightarrow$$ $$y(s)=\frac{y(0)-2x(s)}{s+2}$$

Now we know that $x(0)=1,y(0)=0$ so we get that:

• $$x(s)=\frac{\frac{1}{1+s}+1-2y(s)}{s+5}$$
• $$y(s)=-\frac{2x(s)}{s+2}$$

So, we can substitute them into each other:

• $$x(s)=\frac{\frac{1}{1+s}+1-2\left(-\frac{2x(s)}{s+2}\right)}{s+5}\Longleftrightarrow$$ $$x(s)=\frac{(s+2)^2}{(s+1)^2(s+6)}=\frac{16}{25(s+6)}+\frac{9}{25(s+1)}+\frac{1}{5(s+1)^2}$$
• $$y(s)=-\frac{2\cdot\frac{\frac{1}{1+s}+1-2y(s)}{s+5}}{s+2}\Longleftrightarrow$$ $$y(s)=-\frac{2(s+2)}{(s+1)^2(s+6)}=\frac{8}{25s+6)}-\frac{8}{25(s+1)}-\frac{2}{5(s+1)^2}$$