 # To state: The solution of the given initial value problem using the method of Laplace transforms. Given: The initial value problem is, {y}text{}+{6}{y}'+{5}{y}={12}{e}^{t},{y}{left({0}right)}=-{1},{y}'{left({0}right)}={7} Khaleesi Herbert 2020-12-14 Answered
To state:
The solution of the given initial value problem using the method of Laplace transforms.
Given:
The initial value problem is,
$y+6{y}^{\prime }+5y=12{e}^{t},y\left(0\right)=-1,{y}^{\prime }\left(0\right)=7$
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Approach:
Let The Laplace transformation of f is the function F defined by the integral,
$F\left(s\right)={\int }_{0}^{\mathrm{\infty }}{e}^{-st}f\left(t\right)dt.$
The domain of F (s) is all the values of s for which the integrals exists.
The Laplace transformation of f is denoted by both $F\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathcal{\text{L}}\left\{f\right\}$
To solve initial value problem using Laplace transformation,
a) Take Laplace transformation on both sides of the equation
b) Use the properties of the Laplace transform and the initial conditions to get the equation for the Laplace transform of the solution and then solve the equation for the transform.
c) Determine the inverse Laplace transform of the solution to obtain the final answer.
Calculation:
Take Laplace transform on the both sides of the given initial value problem.
$\mathcal{\text{L}}\left\{y{}^{″}+6{y}^{\prime }+5y\right\}=\mathcal{\text{L}}\left\{12{e}^{t}\right\}$
$\mathcal{\text{L}}\left\{y{}^{″}\right\}\left(s\right)+6\mathcal{\text{L}}\left\{{y}^{\prime }\right\}\left(s\right)+5\mathcal{\text{L}}\left\{y\right\}\left(s\right)=12\mathcal{\text{L}}\left\{{e}^{t}\right\}$
$\left({s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+6\left(sY\left(s\right)-y\left(0\right)\right)+5Y\left(s\right)=12\left(\frac{1}{s-1}\right)$
$\left({s}^{2}Y\left(s\right)-s\left(-1\right)-6\right)+6\left(sY\left(s\right)-\left(-1\right)\right)+5Y\left(s\right)=\frac{12}{s-1}$
$\left({s}^{2}Y\left(s\right)+s-7\right)+6\left(sY\left(s\right)+1\right)+5Y\left(s\right)=\frac{12}{s-1}$
$\left({s}^{2}+6s+5\right)Y\left(s\right)+s-1=\frac{12}{s-1}$
$\left({s}^{2}+6s+5\right)Y\left(s\right)=\frac{12}{s-1}-s$
$Y\left(s\right)=\frac{12}{\left(s-1\right)\left({s}^{2}+6s+5\right)}-\frac{s-1}{{s}^{2}+6s+5}$
$Y\left(s\right)=\frac{12}{\left(s-1\right)\left({s}^{2}+6s+5\right)}-\frac{s-1}{{s}^{2}+6s+5}$
$=\frac{1}{s-1}-\frac{3}{2}\left(\frac{1}{s+1}\right)+\frac{1}{2}\frac{1}{\left(s+5\right)}-\frac{1}{2}\left(\frac{1}{s+1}\right)+\frac{3}{2}\frac{1}{s+5}$
Take Laplac