 # Use the Laplace Transform solve the IVP begin{cases}x'+y=t 4x+y'=0end{cases} x(0)=1 y(0)=2 Harlen Pritchard 2021-02-05 Answered
Use the Laplace Transform solve the IVP
\(\begin{cases}x+y=t\
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it d2saint0
Step 1
Theory :
The Laplace transform of a second order linear differential equation : $y"+\alpha {y}^{\prime }+\beta y=f\left(t\right)$ subject to initial conditions y(0)=A and y'(0)=B is given by,
$Y\left(s\right)=\frac{\left(s+\alpha \right)A+B}{\left({s}^{2}+\alpha s+\beta \right)}+\frac{\left(F\left(s\right)\right)}{\left({s}^{2}+\alpha s+\beta \right)}$
Taking inverse laplace transform on both sides we get, $y\left(t\right)={L}^{-1}\left[\frac{\left(s+\alpha \right)A+B}{\left({s}^{2}+\alpha s+\beta \right)},t\right]+{L}^{-1}\left[\frac{F\left(s\right)}{\left({s}^{2}+\alpha s+\beta \right)},t\right]$
Step 2
Given two simultaneous differential equations with ICs,

Taking Laplace transforms on both sides of each equation and using notation $Y=L\left[y,s\right]$ we obtain,
$sX-x\left(0\right)+Y=L\left[t,s\right]=\frac{1}{{s}^{2}}$
$sY-y\left(0\right)+X=L\left[0,s\right]=0$
Using initial conditions, $sX-1+Y=\frac{1}{{s}^{2}}$
$sY-2+X=0$
$⇒sX+Y=\frac{1}{{s}^{2}}\dots \left(1\right)$
$X+sY=2\dots \left(2\right)$
Solving equations (1) and (2) for X and Y we obtain,
$X=\frac{1-2s}{s\left({s}^{2}-1\right)}\dots \left(3\right)$
$Y=\frac{2s}{{s}^{2}-1}-\frac{1}{{s}^{2}\left({s}^{2}-1\right)}\dots \left(4\right)$
Using partial fractions method in (3) and (4) we further get the values of X and Y as,
$X=-\frac{1}{s}+\frac{s}{\left({s}^{2}-1\right)}-\frac{2}{\left({s}^{2}-1\right)}$
$Y=\frac{5}{4}\cdot \frac{1}{s}+\frac{s}{\left({s}^{2}-1\right)}$
Using inverse laplace transform we get,
$x=-1+\mathrm{cos}ht-\mathrm{sin}ht$
$y=\frac{5}{4}+\mathrm{cos}ht$
The answer is $x=-1+\mathrm{cos}ht-\mathrm{sin}ht$
$y=\frac{5}{4}+\mathrm{cos}ht$