Math / Calculus

Question3) Use Laplace transforms to solve the initial value problem: $y^{\prime \prime}+3 y^{\prime}+2 y=f(t) ; \quad y(0)=0, y^{\prime}(0)=1$ , where $f(t)=\left\{\begin{array}{l}0 \text { if } \mathrm{t}<4 \\ (t-4)^{3} \text { if } t \geq 4\end{array}\right.$ HINT: Do the inversion in two parts like a question on the test.

Studdy Solution

STEP 1

What is this asking? We need to solve a differential equation with an initial value problem using Laplace transforms, where the forcing function $f(t)$ changes its definition at $t=4$ . Watch out! The piecewise function $f(t)$ requires careful handling when applying the Laplace transform.
Don't forget those initial conditions!

STEP 2

1. Rewrite $f(t)$ using the step function
2. Apply the Laplace Transform
3. Solve for $Y(s)$
4. Partial Fraction Decomposition
5. Inverse Laplace Transform

STEP 3

Let's rewrite $f(t)$ using the unit step function, $u(t-4)$ .
This allows us to express the piecewise function in a single expression, which is much easier to work with when doing Laplace transforms.
So, we have $f(t) = (t-4)^3 u(t-4)$ .
This neatly captures the behavior of $f(t)$ , being zero before $t=4$ and $(t-4)^3$ afterwards!

STEP 4

Now, let's apply the Laplace transform to both sides of the differential equation.
Remember the initial conditions $y(0) = 0$ and $y'(0) = 1$ ! $\mathcal{L}\{y''\} + 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{f(t)\}$

STEP 5

Using our handy Laplace transform properties, this becomes: $s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = \mathcal{L}\{(t-4)^3u(t-4)\}$

STEP 6

Substituting the initial conditions and using the time-shift property for the Laplace transform of $f(t)$ , we get: $s^2Y(s) - 1 + 3sY(s) + 2Y(s) = e^{-4s}\frac{6}{s^4}$

STEP 7

Let's solve for $Y(s)$ by factoring and some algebraic manipulation: $Y(s)(s^2 + 3s + 2) = 1 + \frac{6e^{-4s}}{s^4}$ $Y(s) = \frac{1}{s^2 + 3s + 2} + \frac{6e^{-4s}}{s^4(s^2 + 3s + 2)}$ $Y(s) = \frac{1}{(s+1)(s+2)} + \frac{6e^{-4s}}{s^4(s+1)(s+2)}$

STEP 8

Time for partial fraction decomposition!
We'll decompose the two parts of $Y(s)$ separately.
First: $\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$ Solving for $A$ and $B$ gives us $A = 1$ and $B = -1$ .

STEP 9

Now for the second, more involved part: $\frac{6}{s^4(s+1)(s+2)} = \frac{C}{s} + \frac{D}{s^2} + \frac{E}{s^3} + \frac{F}{s^4} + \frac{G}{s+1} + \frac{H}{s+2}$ After a bit of algebra (lots of adding to zero and dividing to one!), we find $C = \frac{3}{2}$ , $D = -3$ , $E = 3$ , $F = 3$ , $G = -2$ , and $H = \frac{1}{2}$ .

STEP 10

Finally, we'll apply the inverse Laplace transform to find $y(t)$ : $y(t) = \mathcal{L}^{-1}\{Y(s)\}$ $y(t) = \mathcal{L}^{-1}\left\{\frac{1}{s+1}\right\} - \mathcal{L}^{-1}\left\{\frac{1}{s+2}\right\} + u(t-4)\mathcal{L}^{-1}\left\{\frac{3/2}{s} - \frac{3}{s^2} + \frac{3}{s^3} + \frac{3}{s^4} - \frac{2}{s+1} + \frac{1/2}{s+2}\right\}\bigg|_{t \to t-4}$

STEP 11

Using the inverse Laplace transform properties, we get our solution: $y(t) = e^{-t} - e^{-2t} + u(t-4)\left(\frac{3}{2} - 3(t-4) + \frac{3}{2}(t-4)^2 + \frac{1}{2}(t-4)^3 - 2e^{-(t-4)} + \frac{1}{2}e^{-2(t-4)}\right)$

STEP 12

$y(t) = e^{-t} - e^{-2t} + u(t-4)\left(\frac{3}{2} - 3(t-4) + \frac{3}{2}(t-4)^2 + \frac{1}{2}(t-4)^3 - 2e^{-(t-4)} + \frac{1}{2}e^{-2(t-4)}\right)$

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Studdy Solution

STEP 1

STEP 2

1. Rewrite $f(t)$ using the step function
2. Apply the Laplace Transform
3. Solve for $Y(s)$
4. Partial Fraction Decomposition
5. Inverse Laplace Transform

STEP 3

STEP 4

Now, let's apply the Laplace transform to both sides of the differential equation.
Remember the initial conditions $y(0) = 0$ and $y'(0) = 1$ ! $\mathcal{L}\{y''\} + 3\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} = \mathcal{L}\{f(t)\}$

STEP 5

Using our handy Laplace transform properties, this becomes: $s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = \mathcal{L}\{(t-4)^3u(t-4)\}$

STEP 6

Substituting the initial conditions and using the time-shift property for the Laplace transform of $f(t)$ , we get: $s^2Y(s) - 1 + 3sY(s) + 2Y(s) = e^{-4s}\frac{6}{s^4}$

STEP 7

Let's solve for $Y(s)$ by factoring and some algebraic manipulation: $Y(s)(s^2 + 3s + 2) = 1 + \frac{6e^{-4s}}{s^4}$ $Y(s) = \frac{1}{s^2 + 3s + 2} + \frac{6e^{-4s}}{s^4(s^2 + 3s + 2)}$ $Y(s) = \frac{1}{(s+1)(s+2)} + \frac{6e^{-4s}}{s^4(s+1)(s+2)}$

STEP 8

Time for partial fraction decomposition!
We'll decompose the two parts of $Y(s)$ separately.
First: $\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$ Solving for $A$ and $B$ gives us $A = 1$ and $B = -1$ .

STEP 9

STEP 10

STEP 11

Using the inverse Laplace transform properties, we get our solution: $y(t) = e^{-t} - e^{-2t} + u(t-4)\left(\frac{3}{2} - 3(t-4) + \frac{3}{2}(t-4)^2 + \frac{1}{2}(t-4)^3 - 2e^{-(t-4)} + \frac{1}{2}e^{-2(t-4)}\right)$

STEP 12

$y(t) = e^{-t} - e^{-2t} + u(t-4)\left(\frac{3}{2} - 3(t-4) + \frac{3}{2}(t-4)^2 + \frac{1}{2}(t-4)^3 - 2e^{-(t-4)} + \frac{1}{2}e^{-2(t-4)}\right)$

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Studdy Solution

STEP 1

STEP 2

1. Rewrite f(t)f(t)f(t) using the step function2. Apply the Laplace Transform3. Solve for Y(s)Y(s)Y(s)4. Partial Fraction Decomposition5. Inverse Laplace Transform

STEP 3

STEP 4

STEP 5

Using our handy Laplace transform properties, this becomes: s2Y(s)−sy(0)−y′(0)+3(sY(s)−y(0))+2Y(s)=L{(t−4)3u(t−4)} s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = \mathcal{L}\{(t-4)^3u(t-4)\} s2Y(s)−sy(0)−y′(0)+3(sY(s)−y(0))+2Y(s)=L{(t−4)3u(t−4)}

STEP 6

Substituting the **initial conditions** and using the time-shift property for the Laplace transform of f(t)f(t)f(t), we get: s2Y(s)−1+3sY(s)+2Y(s)=e−4s6s4 s^2Y(s) - 1 + 3sY(s) + 2Y(s) = e^{-4s}\frac{6}{s^4} s2Y(s)−1+3sY(s)+2Y(s)=e−4ss46​

STEP 7

STEP 8

Time for **partial fraction decomposition**!We'll decompose the two parts of Y(s)Y(s)Y(s) separately.First: 1(s+1)(s+2)=As+1+Bs+2 \frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} (s+1)(s+2)1​=s+1A​+s+2B​ Solving for AAA and BBB gives us A=1A = 1A=1 and B=−1B = -1B=−1.

STEP 9

STEP 10

STEP 11

STEP 12

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1. Rewrite $f(t)$ using the step function
2. Apply the Laplace Transform
3. Solve for $Y(s)$
4. Partial Fraction Decomposition
5. Inverse Laplace Transform

Using our handy Laplace transform properties, this becomes: $s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = \mathcal{L}\{(t-4)^3u(t-4)\}$

Substituting the initial conditions and using the time-shift property for the Laplace transform of $f(t)$ , we get: $s^2Y(s) - 1 + 3sY(s) + 2Y(s) = e^{-4s}\frac{6}{s^4}$

Time for partial fraction decomposition!
We'll decompose the two parts of $Y(s)$ separately.
First: $\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$ Solving for $A$ and $B$ gives us $A = 1$ and $B = -1$ .