use the Laplace transform to solve the given initial-value problem. y"-3y'+2y=4 , y(0)=0 , y'(0)=1

use the Laplace transform to solve the given initial-value problem. y"-3y'+2y=4 , y(0)=0 , y'(0)=1

Question
Laplace transform
asked 2020-11-10
use the Laplace transform to solve the given initial-value problem.
\(y"-3y'+2y=4 \ , \ y(0)=0 \ , \ y'(0)=1\)

Answers (1)

2020-11-11
\(\text{Step 1 }\)
\(\text{Given differential equation is}\)
\(y"-3y'+2y=4 \ , \ y(0)=0 \ , \ y'(0)=1\)
\(\text{Step 2}\)
\(\text{Laplace transform is denoted as}\)
\(Y(s)=L(y(t))\)
\(\text{According the Laplace formula, }\)
\(L(y^{n}(t))=s^{n}Y(s)-s^{n-1}y(0)-s^{n-2}y'(0)-\dotsb-y^{n-1}(0)\)
\(\text{Let's take the Laplace on both sides, of given differential equation}\)
\(L(y"-3y'+2y)=L(4)\)
\(s^2Y(s)-sy(0)-y'(0)-3\left[sY(s)-y(0)\right]+2Y(s)=\frac{4}{s}\)
\(Y(s)(s^2-3s+2)+y(0)(3-s)-y'(0)=\frac{4}{s}\)
\(\text{By initial conditions, }\)
\(Y(s)(s^2-3s+2)-1-\frac{4}{s}=0\)
\(Y(s)=\frac{\frac{4+s}{s}}{s^2-3s+2}\)
\(=\frac{4+s}{s(s-2)(s-1)}\)
\(\text{By partial fraction}\)
\(Y(s)=\frac{2}{s}-\frac{5}{s-1}+\frac{3}{s-2}\)
\(\text{Taking the inverse Laplace, }\)
\(y(t)=2-5e^{t}+3^{2t}\)
\(\text{This is the required general solution.}\)
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