Use Laplace transforms to find the solution to the following set of equations

$\frac{d y_{1}}{d t}+y_{2}=x$

$\frac{d y_{2}}{d t}+3 y_{2}=2 y_{1}$

for $x=e^{-t}$ and zero initial conditions.

Question

Use Laplace transforms to find the solution to the following set of equations

[latex]\frac{d y_{1}}{d t}+y_{2}=x[/latex]

[latex]\frac{d y_{2}}{d t}+3 y_{2}=2 y_{1}[/latex]

for [latex]x=e^{-t}[/latex] and zero initial conditions.

Accepted Answer

Laplace transform of the system of ODEs gives:

[latex]L \left(\frac{d y_{1}}{d t}\right)+ L \left(y_{2}\right)= L \left(e^{-t}\right)[/latex]

[latex]L \left(\frac{d y_{2}}{d t}\right)+3 L \left(y_{2}\right)=2 L \left(y_{1}\right)[/latex]

[latex]s Y_{1}+Y_{2}=\frac{1}{s+1}[/latex] (1)

[latex]s Y_{2}+3 Y_{2}=2 Y_{1}[/latex] (2)

Next solve Equation 2 for [latex]Y_{2}[/latex] in terms of [latex]Y_{1}[/latex]

[latex]Y_{2}(s+3)=2 Y_{1}[/latex]

[latex]Y_{2}=\frac{2 Y_{1}}{s+3}[/latex] (3)

Substitute equation 3 into equation 1 and solve for [latex]Y_{1}[/latex]

[latex]s Y_{1}+\frac{2 Y_{1}}{s+3}=\frac{1}{s+1}[/latex]

[latex]Y_{1}\left(s+\frac{2}{s+3}\right)=\frac{1}{s+1}[/latex]

[latex]Y_{1}=\frac{1}{(s+1)\left(s+\frac{2}{s+3}\right)}[/latex]

Expand using partial fractions:

[latex]Y_{1}=\frac{s+3}{(s+1)^{2}(s+2)}=\frac{1}{s+2}-\frac{1}{s+1}+\frac{2}{(s+1)^{2}}[/latex]

Now go back and substitute into equation 3 to get [latex]Y_{2}[/latex] and expand using partial fractions:

[latex]Y_{2}=\frac{2 Y_{1}}{s+3}=\frac{2}{(s+1)^{2}(s+2)}=\frac{2}{s+2}-\frac{2}{s+1}+\frac{2}{(s+1)^{2}}[/latex]

Finally, get both time-domain solutions using the inverse Laplace transform:

[latex]y_{1}(t)=e^{-2 t}-e^{-t}+2 t e^{-t}[/latex]

[latex]y_{2}(t)=2\left(e^{-2 t}-e^{-t}+t e^{-t}\right)[/latex]

Question 3.17: Use Laplace transforms to find the solution to the following...