Math / Calculus

Question(1 point)
Consider the initial value problem $y^{\prime}+2 y=8 t, \quad y(0)=6$ a. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of $y(t)$ by $Y(s)$ . Do not move any terms from one side of the equation to the other (until you get to part (b) below).
$\square$ help (formulas) b. Solve your equation for $Y(s)$ . $Y(\delta)=\mathcal{L}\{y(t)\}=$ $\square$ c. Take the inverse Laplace transform of both sides of the previous equation to solve for $y(t)$ . $y(t)=\square \text { (D) }$

Studdy Solution

STEP 1

What is this asking? We're given a differential equation with an initial condition and we need to solve it using the Laplace transform. Watch out! Don't mix up the Laplace transform formulas and be careful with algebraic manipulations when solving for $Y(s)$ .

STEP 2

1. Laplace Transform of the Differential Equation
2. Solve for Y(s)
3. Inverse Laplace Transform

STEP 3

Let's transform both sides of our differential equation $y^{\prime} + 2y = 8t$ .
Remember, we denote the Laplace transform of $y(t)$ as $Y(s)$ .
Using our handy-dandy Laplace transform formulas, the transform of $y^{\prime}$ is $sY(s) - y(0)$ , and the transform of $2y$ is $2Y(s)$ .
The transform of $8t$ is $\frac{8}{s^2}$ .

STEP 4

We know that $y(0) = \mathbf{6}$ , so let's plug that in.
Our transformed equation becomes $sY(s) - \mathbf{6} + 2Y(s) = \frac{8}{s^2}$ .

STEP 5

We want to solve for $Y(s)$ , so let's factor it out on the left side: $(s + 2)Y(s) - 6 = \frac{8}{s^2}$ .

STEP 6

Now, let's isolate $Y(s)$ .
First, add $6$ to both sides: $(s + 2)Y(s) = \frac{8}{s^2} + 6$ .
Then, divide both sides by $(s + 2)$ : $Y(s) = \frac{\frac{8}{s^2} + 6}{s + 2}$ .

STEP 7

Let's simplify that expression for $Y(s)$ a bit.
We can rewrite $6$ as $\frac{6s^2}{s^2}$ to get a common denominator: $Y(s) = \frac{\frac{8 + 6s^2}{s^2}}{s + 2}$ .
This simplifies to $Y(s) = \frac{8 + 6s^2}{s^2(s + 2)}$ .

STEP 8

To make it easier to find the inverse Laplace transform, let's do a partial fraction decomposition of $\frac{6s^2 + 8}{s^2(s + 2)}$ .
We're looking for something of the form $\frac{A}{s} + \frac{B}{s^2} + \frac{C}{s+2}$ .
Solving for A, B, and C, we find $A = -1$ , $B = 4$ , and $C = 7$ .
So, we have $Y(s) = \frac{-1}{s} + \frac{4}{s^2} + \frac{7}{s + 2}$ .

STEP 9

Now, we can apply the inverse Laplace transform to each term!
The inverse transform of $\frac{-1}{s}$ is $-1$ , the inverse transform of $\frac{4}{s^2}$ is $4t$ , and the inverse transform of $\frac{7}{s + 2}$ is $7e^{-2t}$ .

STEP 10

Putting it all together, we get $y(t) = -1 + 4t + 7e^{-2t}$ .

STEP 11

$y(t) = -1 + 4t + 7e^{-2t}$

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Studdy Solution

STEP 1

What is this asking? We're given a differential equation with an initial condition and we need to solve it using the Laplace transform. Watch out! Don't mix up the Laplace transform formulas and be careful with algebraic manipulations when solving for $Y(s)$ .

STEP 2

1. Laplace Transform of the Differential Equation
2. Solve for Y(s)
3. Inverse Laplace Transform

STEP 3

STEP 4

We know that $y(0) = \mathbf{6}$ , so let's plug that in.
Our transformed equation becomes $sY(s) - \mathbf{6} + 2Y(s) = \frac{8}{s^2}$ .

STEP 5

We want to solve for $Y(s)$ , so let's factor it out on the left side: $(s + 2)Y(s) - 6 = \frac{8}{s^2}$ .

STEP 6

Now, let's isolate $Y(s)$ .
First, add $6$ to both sides: $(s + 2)Y(s) = \frac{8}{s^2} + 6$ .
Then, divide both sides by $(s + 2)$ : $Y(s) = \frac{\frac{8}{s^2} + 6}{s + 2}$ .

STEP 7

Let's simplify that expression for $Y(s)$ a bit.
We can rewrite $6$ as $\frac{6s^2}{s^2}$ to get a common denominator: $Y(s) = \frac{\frac{8 + 6s^2}{s^2}}{s + 2}$ .
This simplifies to $Y(s) = \frac{8 + 6s^2}{s^2(s + 2)}$ .

STEP 8

STEP 9

Now, we can apply the inverse Laplace transform to each term!
The inverse transform of $\frac{-1}{s}$ is $-1$ , the inverse transform of $\frac{4}{s^2}$ is $4t$ , and the inverse transform of $\frac{7}{s + 2}$ is $7e^{-2t}$ .

STEP 10

Putting it all together, we get $y(t) = -1 + 4t + 7e^{-2t}$ .

STEP 11

$y(t) = -1 + 4t + 7e^{-2t}$

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Studdy Solution

STEP 1

What is this asking? We're given a *differential equation* with an *initial condition* and we need to solve it using the *Laplace transform*. Watch out! Don't mix up the *Laplace transform* formulas and be careful with algebraic manipulations when solving for Y(s)Y(s)Y(s).

STEP 2

1. Laplace Transform of the Differential Equation2. Solve for Y(s)3. Inverse Laplace Transform

STEP 3

STEP 4

We know that y(0)=6y(0) = \mathbf{6}y(0)=6, so let's plug that in.Our transformed equation becomes sY(s)−6+2Y(s)=8s2sY(s) - \mathbf{6} + 2Y(s) = \frac{8}{s^2}sY(s)−6+2Y(s)=s28​.

STEP 5

We want to solve for Y(s)Y(s)Y(s), so let's **factor** it out on the left side: (s+2)Y(s)−6=8s2(s + 2)Y(s) - 6 = \frac{8}{s^2}(s+2)Y(s)−6=s28​.

STEP 6

Now, let's **isolate** Y(s)Y(s)Y(s).First, **add** 666 to both sides: (s+2)Y(s)=8s2+6(s + 2)Y(s) = \frac{8}{s^2} + 6(s+2)Y(s)=s28​+6.Then, **divide** both sides by (s+2)(s + 2)(s+2): Y(s)=8s2+6s+2Y(s) = \frac{\frac{8}{s^2} + 6}{s + 2}Y(s)=s+2s28​+6​.

STEP 7

STEP 8

STEP 9

Now, we can **apply** the inverse Laplace transform to each term!The inverse transform of −1s\frac{-1}{s}s−1​ is −1-1−1, the inverse transform of 4s2\frac{4}{s^2}s24​ is 4t4t4t, and the inverse transform of 7s+2\frac{7}{s + 2}s+27​ is 7e−2t7e^{-2t}7e−2t.

STEP 10

Putting it all together, we get y(t)=−1+4t+7e−2ty(t) = -1 + 4t + 7e^{-2t}y(t)=−1+4t+7e−2t.

STEP 11

y(t)=−1+4t+7e−2ty(t) = -1 + 4t + 7e^{-2t}y(t)=−1+4t+7e−2t

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What is this asking? We're given a differential equation with an initial condition and we need to solve it using the Laplace transform. Watch out! Don't mix up the Laplace transform formulas and be careful with algebraic manipulations when solving for $Y(s)$ .

1. Laplace Transform of the Differential Equation
2. Solve for Y(s)
3. Inverse Laplace Transform

We know that $y(0) = \mathbf{6}$ , so let's plug that in.
Our transformed equation becomes $sY(s) - \mathbf{6} + 2Y(s) = \frac{8}{s^2}$ .

We want to solve for $Y(s)$ , so let's factor it out on the left side: $(s + 2)Y(s) - 6 = \frac{8}{s^2}$ .

Now, let's isolate $Y(s)$ .
First, add $6$ to both sides: $(s + 2)Y(s) = \frac{8}{s^2} + 6$ .
Then, divide both sides by $(s + 2)$ : $Y(s) = \frac{\frac{8}{s^2} + 6}{s + 2}$ .

Now, we can apply the inverse Laplace transform to each term!
The inverse transform of $\frac{-1}{s}$ is $-1$ , the inverse transform of $\frac{4}{s^2}$ is $4t$ , and the inverse transform of $\frac{7}{s + 2}$ is $7e^{-2t}$ .

Putting it all together, we get $y(t) = -1 + 4t + 7e^{-2t}$ .

$y(t) = -1 + 4t + 7e^{-2t}$