To state:The solution of the given initial value problem using the method of Laplace transforms.Given:The initial value problem is,y+6y′+5y=12et,y(0)=−1,y′(0)=7

Question

firmablogF · Accepted Answer

Approach:Let&amp;nbsp;f(t)&amp;nbsp;be a function on&amp;nbsp;[0,∞).&amp;nbsp;The Laplace transformation of f is the function F defined by the integral,F(s)=∫0∞e−stf(t)&amp;nbsp;dt&amp;nbsp;.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&amp;nbsp;F&amp;nbsp;and&amp;nbsp;L{f}Using the Laplace transformation, the initial value problem can be solved.a) Apply the Laplace transformation to the entire equation.b) Determine the equation for the Laplace transform of the solution using the initial conditions and the Laplace transform&#039;s characteristics, and then work out 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.L{y&amp;nbsp;″+6y′+5y}=L{12et}L{y&amp;nbsp;″}(s)+6L{y′}(s)+5L{y}(s)=12L{et}(s2Y(s)−sy(0)−y′(0))+6(sY(s)−y(0))+5Y(s)=12(1s−1)(s2Y(s)−s(−1)−6)+6(sY(s)−(−1))+5Y(s)=12s−1(s2Y(s)+s−7)+6(sY(s)+1)+5Y(s)=12s−1(s2+6s+5)Y(s)+s−1=12s−1(s2+6s+5)Y(s)=12s−1−sY(s)=12(s−1)(s2+6s+5)−s−1s2+6s+5Y(s)=12(s−1)(s2+6s+5)−s−1s2+6s+5=1s−1−32(1s+1)+121(s+5)−12(1s+1)+321s+5Take Laplac

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}

Answered question

Answer & Explanation

New Questions in Integral Calculus