geduiwelh

2021-02-13

Use a Laplace transform to determine the solution of the following systems with differential equations

a)${x}^{\prime}+4x+3y=0\text{with}x(0)=0$

${y}^{\prime}+3x+4y=2{e}^{t},y(0)=0$

a)

Brighton

Skilled2021-02-14Added 103 answers

Solution:

The system of differential equations is

${x}^{\prime}+4x+3y=0$

${y}^{\prime}+3x+4y=2{e}^{t}$

Apply Laplace transform:

$sX(s)-x(0)+4X(s)+3Y(s)=0\Rightarrow (s+4)X(s)+3Y(s)=0$

$sY(s)-y(0)+3X(s)+4Y(s)=2\left(\frac{1}{(s-1)}\right)\Rightarrow 3X(s)+(s+4)Y(s)=\frac{2}{(s-1)}$

Perform substitution:

$Y(s)=-\frac{(s+4)}{3}X(s),\text{so}3X(s)-\frac{(s+4{)}^{2}}{3}X(s)=\frac{2}{(s-1)}$

$X(s)=\frac{6}{(9-(s+4{)}^{2})(s-1)},\text{so}Y(s)=-\frac{2(s+4)}{(9-(s+4{)}^{2})(s-1)}$

$X(s)=-\frac{6}{(s+7)(s+1)(s-1)},\text{so}Y(s)=\frac{2(s+4)}{(s+7)(s+1)(s-1)}$

Apply inverse Laplace transform:

Perform partial fractional decomposition:

$X(s)=-\frac{6}{(s+7)(s+1)(s-1)}=-\frac{1}{8(s+7)}+\frac{1}{2(s+1)}-\frac{3}{8(s-1)}$

Apply inverse Laplace transform:

$x(t)=-\frac{1}{8}{e}^{-7t}+\frac{1}{2}{e}^{-t}-\frac{3}{8}{e}^{t}$

Perform partial fractional decomposition:

$Y(s)=\frac{2(s+4)}{(s+7)(s+1)(s-1)}=-\frac{1}{8(s+7)}-\frac{1}{2(s+1)}+\frac{5}{8(s-1)}$

Apply inverse Laplace transform:

$y(t)=-\frac{1}{8}{e}^{-7t}-\frac{1}{2}{e}^{-t}+\frac{5}{8}{e}^{t}$

The system of differential equations is

Apply Laplace transform:

Perform substitution:

Apply inverse Laplace transform:

Perform partial fractional decomposition:

Apply inverse Laplace transform:

Perform partial fractional decomposition:

Apply inverse Laplace transform:

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