# Use Laplace transform to find the solution of the IVP2y'+y=0 , y(0)=-3a) f{{left({t}right)}}

Use Laplace transform to find the solution of the IVP
$2{y}^{\prime }+y=0,y\left(0\right)=-3$
a) $f\left(t\right)=3{e}^{-2t}$
b)$f\left(t\right)=3{e}^{\frac{t}{2}}$
c)$f\left(t\right)=6{e}^{2t}$
d) $f\left(t\right)=3{e}^{-\frac{t}{2}}$

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Step 1
Given
Use Laplace transform to find the solution of the IVP,
$2{y}^{\prime }+y=0,y\left(0\right)=-3$
step 2
Solution
$2{y}^{\prime }+y=0$
Taking Laplace transform on both sides
$L\left(2{y}^{\prime }\right)+L\left(y\right)=0$
$2L\left({y}^{\prime }\right)+L\left(y\right)=0$
$2\left[\mathrm{\Delta }L\left(y\right)-y\left(0\right)\right]+L\left(y\right)=0$
$2\mathrm{\Delta }L\left(y\right)-2y\left(0\right)+L\left(y\right)=0$
$2\mathrm{\Delta }L\left(y\right)-2\left(-3\right)+L\left(y\right)=0$
$L\left(y\right)\left[2\mathrm{\Delta }+1\right]=-6$
$L\left(y\right)=\frac{-6}{2\mathrm{\Delta }+1}$
$=\frac{-6}{2\left(\mathrm{\Delta }+\frac{1}{2}\right)}$
$Y={L}^{-1}\left(\frac{-3}{\mathrm{\Delta }+\frac{1}{2}}\right)=-3{L}^{-1}\left(\frac{1}{\mathrm{\Delta }+\frac{1}{2}}\right)$
$Y\left(t\right)=-3{e}^{-\frac{t}{2}}$
Therefore the required answer is $f\left(t\right)=-3{e}^{-\frac{t}{2}}$