Problem 7. Using Laplace transforms, find the solution y(t)...

Problem 7. Using Laplace transforms, find the solution $y(t)$ to the linear ordinary differential equation $\ddot{y}(t) + 7\dot{y}(t) + 12y(t) = \dot{u}(t) - u(t)$, where $u(t) = e^{-t}$, $\dot{y}(0) = 0$, and $y(0) = 0$. Note that this is the forced response because the initial conditions are zero.

Transcript

00:01 The question is to solve the differential equation y double dash of t plus 7 y dash of t plus 6 y of t is equal to delta t minus 3 where y of 0 is equal to 0 and y dash of 0 is equal to 0 using leplice transforms.

00:21 If the leplice transform l of a function f of t is capital f of s then the leplice transform, then the leplice transform.

00:30 The laplace transform of the nth derivative, fn of t, is defined as s raise to n, f of s minus s, minus to n minus 1, f of 0, minus, et cetera, minus fn minus 1 of 0.

00:48 So here, let the laplace transform of y of t be capital y of s, then the laplice transform of y, d, be capital y of s, then the laplice transform of y double dash of is s square y of s minus s times y of 0 minus y dash of 0.

01:13 Since y of 0 and y dash of 0 is 0, we have this as s square y of s.

01:19 Similarly, they'll apply transform of y dash of t using the above formula is s y of s minus y of 0.

01:28 Since y of 0 it is s y of s y of s.

01:31 Also the laplice transform of delta of t minus a is e raised to negative as so that the laplice transform of delta of t minus 3 is e raised to negative 3s so that the equation this can be converted using laplice transform by taking laplice transform on both sides and applying the linearity of laplice transforms to get the equation as.

02:01 S square y of s plus 7 s y of s plus 7 s plus 6 y of s is equal to e raised to negative 3 s.

02:11 From this we can solve for y of s as y of s is equal to erase to negative 3 s divided by s 2 plus 7 s plus 6...

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