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

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

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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