geduiwelh

2021-02-13

Use a Laplace transform to determine the solution of the following systems with differential equations
a)
${y}^{\prime }+3x+4y=2{e}^{t},y\left(0\right)=0$

Brighton

Solution:
The system of differential equations is
${x}^{\prime }+4x+3y=0$
${y}^{\prime }+3x+4y=2{e}^{t}$
Apply Laplace transform:
$sX\left(s\right)-x\left(0\right)+4X\left(s\right)+3Y\left(s\right)=0⇒\left(s+4\right)X\left(s\right)+3Y\left(s\right)=0$
$sY\left(s\right)-y\left(0\right)+3X\left(s\right)+4Y\left(s\right)=2\left(\frac{1}{\left(s-1\right)}\right)⇒3X\left(s\right)+\left(s+4\right)Y\left(s\right)=\frac{2}{\left(s-1\right)}$
Perform substitution:

Apply inverse Laplace transform:
Perform partial fractional decomposition:
$X\left(s\right)=-\frac{6}{\left(s+7\right)\left(s+1\right)\left(s-1\right)}=-\frac{1}{8\left(s+7\right)}+\frac{1}{2\left(s+1\right)}-\frac{3}{8\left(s-1\right)}$
Apply inverse Laplace transform:
$x\left(t\right)=-\frac{1}{8}{e}^{-7t}+\frac{1}{2}{e}^{-t}-\frac{3}{8}{e}^{t}$
Perform partial fractional decomposition:
$Y\left(s\right)=\frac{2\left(s+4\right)}{\left(s+7\right)\left(s+1\right)\left(s-1\right)}=-\frac{1}{8\left(s+7\right)}-\frac{1}{2\left(s+1\right)}+\frac{5}{8\left(s-1\right)}$
Apply inverse Laplace transform:
$y\left(t\right)=-\frac{1}{8}{e}^{-7t}-\frac{1}{2}{e}^{-t}+\frac{5}{8}{e}^{t}$

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