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

Laplace transform

Use Laplace transform to find the solution of the IVP
$$2y'+y=0 , y(0)=-3$$
a) $$f{{\left({t}\right)}}={3}{e}^{{-{2}{t}}}$$
b)$$f{{\left({t}\right)}}={3}{e}^{{\frac{t}{{2}}}}$$
c)$$f{{\left({t}\right)}}={6}{e}^{{{2}{t}}}$$
d) $$f{{\left({t}\right)}}={3}{e}^{{-\frac{t}{{2}}}}$$

2021-01-31

Step 1
Given
Use Laplace transform to find the solution of the IVP,
$$2y'+y=0, y(0)=-3$$
step 2
Solution
$$2y'+y=0$$
Taking Laplace transform on both sides
$$L(2y')+L(y)=0$$
$$2L(y')+L(y)=0$$
$${2}{\left[\Delta{L}{\left({y}\right)}-{y}{\left({0}\right)}\right]}+{L}{\left({y}\right)}={0}$$
$${2}\Delta{L}{\left({y}\right)}-{2}{y}{\left({0}\right)}+{L}{\left({y}\right)}={0}$$
$${2}\Delta{L}{\left({y}\right)}-{2}{\left(-{3}\right)}+{L}{\left({y}\right)}={0}$$
$${L}{\left({y}\right)}{\left[{2}\Delta+{1}\right]}=-{6}$$
$${L}{\left({y}\right)}=\frac{{-{6}}}{{{2}\Delta+{1}}}$$
$$=\frac{{-{6}}}{{{2}{\left(\Delta+\frac{1}{{2}}\right)}}}$$
$${Y}={L}^{ -{{1}}}{\left(\frac{{-{3}}}{{\Delta+\frac{1}{{2}}}}\right)}=-{3}{L}^{ -{{1}}}{\left(\frac{{{1}}}{{\Delta+\frac{1}{{2}}}}\right)}$$
$${Y}{\left({t}\right)}=-{3}{e}^{{-\frac{t}{{2}}}}$$
Therefore the required answer is $$f{{\left({t}\right)}}=-{3}{e}^{{-\frac{t}{{2}}}}$$