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Math  /  Calculus

QuestionTake the Laplace transform of the following initial value problem and solve for Y(s)=L{y(t)}Y(s)=\mathcal{L}\{y(t)\} : Y(s)=□.y′′−4y′−21y={1,0≤t<10,1≤ty(0)=0,y′(0)=0Y(s)=\square . \quad y^{\prime \prime}-4 y^{\prime}-21 y=\left\{\begin{array}{ll} 1, & 0 \leq t<1 \\ 0, & 1 \leq t \end{array} \quad y(0)=0, y^{\prime}(0)=0\right. □\square Now find the inverse transform to find y(t)=y(t)= □\square (Use step(t-c) for uc(t)u_{c}(t) ). Note: 1s(s−7)(s+3)=−121s+130s+3+170s−7\frac{1}{s(s-7)(s+3)}=\frac{-\frac{1}{21}}{s}+\frac{\frac{1}{30}}{s+3}+\frac{\frac{1}{70}}{s-7}

Studdy Solution

STEP 1

What is this asking? We need to solve a *differential equation* with an interesting right-hand side that changes at t=1t = 1, using the Laplace transform, and then transform back to get y(t)y(t). Watch out! Remember to correctly represent the piecewise function on the right-hand side when taking the Laplace transform!
Also, don't forget about those initial conditions.

STEP 2

1. Laplace Transform
2. Partial Fraction Decomposition
3. Inverse Laplace Transform

STEP 3

Alright, let's **kick things off** by taking the Laplace transform of our differential equation.
The left-hand side is pretty straightforward.
Remember, L{y′′(t)}=s2Y(s)−sy(0)−y′(0)\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) and L{y′(t)}=sY(s)−y(0)\mathcal{L}\{y'(t)\} = sY(s) - y(0).
Since y(0)=0y(0) = 0 and y′(0)=0y'(0) = 0, things simplify nicely!

STEP 4

So, the Laplace transform of the left-hand side becomes s2Y(s)−4sY(s)−21Y(s)s^2Y(s) - 4sY(s) - 21Y(s), which we can write as (s2−4s−21)Y(s)(s^2 - 4s - 21)Y(s).

STEP 5

Now, for the right-hand side, we've got that piecewise function.
We can rewrite it using the unit step function as 1−u1(t)1 - u_1(t).
The Laplace transform of 1 is 1s\frac{1}{s}, and the Laplace transform of u1(t)u_1(t) is e−ss\frac{e^{-s}}{s}.
So, the Laplace transform of the right-hand side is 1s−e−ss\frac{1}{s} - \frac{e^{-s}}{s}, which simplifies to 1−e−ss\frac{1 - e^{-s}}{s}.

STEP 6

Putting it all together, our transformed equation is (s2−4s−21)Y(s)=1−e−ss(s^2 - 4s - 21)Y(s) = \frac{1 - e^{-s}}{s}.
Now, we can **isolate** Y(s)Y(s) to get Y(s)=1−e−ss(s2−4s−21)Y(s) = \frac{1 - e^{-s}}{s(s^2 - 4s - 21)}.
Factoring the denominator gives us Y(s)=1−e−ss(s−7)(s+3)Y(s) = \frac{1 - e^{-s}}{s(s-7)(s+3)}.

STEP 7

The problem conveniently gives us the partial fraction decomposition of 1s(s−7)(s+3)\frac{1}{s(s-7)(s+3)}, which is −1211s+1301s+3+1701s−7-\frac{1}{21}\frac{1}{s} + \frac{1}{30}\frac{1}{s+3} + \frac{1}{70}\frac{1}{s-7}.
So, we can **substitute** this back into our expression for Y(s)Y(s) to get Y(s)=(1−e−s)(−1211s+1301s+3+1701s−7)Y(s) = (1 - e^{-s})\left(-\frac{1}{21}\frac{1}{s} + \frac{1}{30}\frac{1}{s+3} + \frac{1}{70}\frac{1}{s-7}\right).

STEP 8

Now, for the **grand finale**, we'll take the inverse Laplace transform!
Remember, the inverse Laplace transform is linear, so we can take it term by term.

STEP 9

The inverse Laplace transform of 1s\frac{1}{s} is 1, the inverse Laplace transform of 1s+3\frac{1}{s+3} is e−3te^{-3t}, and the inverse Laplace transform of 1s−7\frac{1}{s-7} is e7te^{7t}.

STEP 10

Also, recall that the inverse Laplace transform of e−csF(s)e^{-cs}F(s) is uc(t)f(t−c)u_c(t)f(t-c), where f(t)f(t) is the inverse Laplace transform of F(s)F(s).

STEP 11

Putting it all together, we get y(t)=−121+130e−3t+170e7t−u1(t)(−121+130e−3(t−1)+170e7(t−1))y(t) = -\frac{1}{21} + \frac{1}{30}e^{-3t} + \frac{1}{70}e^{7t} - u_1(t)\left(-\frac{1}{21} + \frac{1}{30}e^{-3(t-1)} + \frac{1}{70}e^{7(t-1)}\right).

STEP 12

y(t)=−121+130e−3t+170e7t−u1(t)(−121+130e−3(t−1)+170e7(t−1))y(t) = -\frac{1}{21} + \frac{1}{30}e^{-3t} + \frac{1}{70}e^{7t} - u_1(t)\left(-\frac{1}{21} + \frac{1}{30}e^{-3(t-1)} + \frac{1}{70}e^{7(t-1)}\right)

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