Axsah Rachel Punnen

2022-07-07

Solve the ODE by Laplace transform

y^2-3y^1+2y=4t-8; y(0)=2,

xleb123

To solve the given ordinary differential equation (ODE) using Laplace transform, we will transform the equation into the Laplace domain, solve for the Laplace transform of the unknown function, and then use inverse Laplace transform to obtain the solution in the time domain.
The given ODE is:
${y}^{2}-3{y}^{1}+2y=4t-8$
Step 1: Taking the Laplace transform of both sides of the equation.
Applying the Laplace transform to each term, we get:
$ℒ\left\{{y}^{2}-3{y}^{1}+2y\right\}=ℒ\left\{4t-8\right\}$
Using the linearity property of the Laplace transform, we can split the left-hand side as follows:
$ℒ\left\{{y}^{2}\right\}-3ℒ\left\{{y}^{1}\right\}+2ℒ\left\{y\right\}=ℒ\left\{4t-8\right\}$
Step 2: Applying the Laplace transform to each term.
Using the derivative property of the Laplace transform, we have:
$ℒ\left\{{y}^{2}\right\}={s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$ℒ\left\{{y}^{1}\right\}=sY\left(s\right)-y\left(0\right)$
$ℒ\left\{y\right\}=Y\left(s\right)$
Here, $Y\left(s\right)$ represents the Laplace transform of the unknown function $y\left(t\right)$.
Step 3: Substituting the transformed terms back into the equation.
After substitution, the equation becomes:
${s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)-3\left(sY\left(s\right)-y\left(0\right)\right)+2Y\left(s\right)=\frac{4}{s}-\frac{8}{s}$
Simplifying the equation further, we have:
$\left({s}^{2}-3s+2\right)Y\left(s\right)-sy\left(0\right)+{y}^{\prime }\left(0\right)-3y\left(0\right)=\frac{4}{s}-\frac{8}{s}$
Step 4: Substituting the initial conditions.
Given initial conditions are:
$y\left(0\right)=2$
Substituting this value into the equation, we get:
$\left({s}^{2}-3s+2\right)Y\left(s\right)-2s+{y}^{\prime }\left(0\right)-6=\frac{4}{s}-\frac{8}{s}$
Step 5: Solving for $Y\left(s\right)$.
Rearranging the equation to solve for $Y\left(s\right)$, we have:
$\left({s}^{2}-3s+2\right)Y\left(s\right)=\frac{4}{s}-\frac{8}{s}+2s+6$
$\left({s}^{2}-3s+2\right)Y\left(s\right)=\frac{4-8s}{s}+2s+6$
$\left({s}^{2}-3s+2\right)Y\left(s\right)=\frac{4-8s+2{s}^{2}+6s}{s}$
$\left({s}^{2}-3s+2\right)Y\left(s\right)=\frac{2{s}^{2}-2s+4}{s}$
Step 6: Simplifying the right-hand side.
Factoring the numerator of the right-hand side, we get:
$\left({s}^{2}-3s+2\right)Y\left(s\right)=\frac{2\left({s}^{2}-s+2\right)}{s}$
Step 7: Canceling common factors.
Canceling out the common factor $\left({s}^{2}-s+2\right)$, we have:
$Y\left(s\right)=\frac{2}{s}$
Step 8: Taking the inverse Laplace transform.
To find the solution $y\left(t\right)$, we need to take the inverse Laplace transform of $Y\left(s\right)$.
${ℒ}^{-1}\left\{Y\left(s\right)\right\}={ℒ}^{-1}\left\{\frac{2}{s}\right\}$
Using the inverse transform property, we obtain:
$y\left(t\right)=2$
Therefore, the solution to the given ODE with the initial condition $y\left(0\right)=2$ is $y\left(t\right)=2$.

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