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2. (60p) Solve the given initial value problem using LAPLACE TRANSFORMS. \* The solutions with different methods won't be accepted. y'' - 2y' + 2y = e$^{-t}$ y(0) = 0, y'(0) = 1 Note: Your solutions should be written in detail. (Partial fractions, Laplace transforms) TABLE 6.2.1 Elementary Laplace Transforms f(t) = \mathcal{L}^{-1} \{F(s)\} $e^{at}$ $e^{at}sin(bt)$ $e^{at}cos(bt)$ F(s) = \mathcal{L} \{f(t)\} $\frac{1}{s - a}$, $s > a$ $\frac{b}{(s - a)^2 + b^2}$, $s > a$ $\frac{s - a}{(s - a)^2 + b^2}$, $s > a$

          2. (60p) Solve the given initial value problem using LAPLACE TRANSFORMS.
\* The solutions with different methods won't be accepted.
y'' - 2y' + 2y = e$^{-t}$  y(0) = 0, y'(0) = 1
Note: Your solutions should be written in detail. (Partial fractions, Laplace
transforms)
TABLE 6.2.1 Elementary Laplace Transforms
f(t) = \mathcal{L}^{-1} \{F(s)\}
$e^{at}$
$e^{at}sin(bt)$
$e^{at}cos(bt)$
F(s) = \mathcal{L} \{f(t)\}
$\frac{1}{s - a}$, $s > a$
$\frac{b}{(s - a)^2 + b^2}$, $s > a$
$\frac{s - a}{(s - a)^2 + b^2}$, $s > a$

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$2. (60p) Solve the given initial value problem using LAPLACE TRANSFORMS. The solutions with different methods won't be accepted. y” - 2y' + 2y = e^-t y(0) = 0, y'(0) = 1 Note: Your solutions should be written in detail. (Partial fractions, Laplace transforms) TABLE 6.2.1 Elementary Laplace Transforms f(t) = ℒ^-1 {F(s)} e^at e^atsin(bt) e^atcos(bt) F(s) = ℒ {f(t)} (1)/(s - a), s > a (b)/((s - a)^2 + b^2), s > a (s - a)/((s - a)^2 + b^2), s > a$

Added by Brianna B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Madhur L

08/08/2023

Step 1

The Laplace transform of y''(t) is s^2Y(s) - sy(0) - y'(0). The Laplace transform of y'(t) is sY(s) - y(0). The Laplace transform of y(t) is Y(s). Using these formulas, we can rewrite the differential equation as: s^2Y(s) - sy(0) - y'(0) - 2(sY(s) - y(0)) + Show more…

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Please kindly solve within 30 minutes. It's very urgent. Thank you. 2. (6Op) Solve the given initial value problem using LAPLACE TRANSFORMS. The solutions with different methods won't be accepted. y"-2y'+2y=e-t, y0=0, y0=1 Note: Your solutions should be written in detail. (Partial fractions, Laplace transforms) TABLE 6.2.1 Elementary Laplace Transforms f=-1{Fs} Fs={f} ear s > a eat sin(bt) b s > a 29 + 20 - s s - a s > a s - a + b^2 eat cos(bt)

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Transcript

00:01 Hello everyone, welcome to this video.

00:03 Here we are given with the initial value problem that is y double dash 6y dash plus 8y minus 3 cos t which is equal to minus 3 cos t minus 2 sin t with the condition y of pi by 2 is equal to 1 and y dash of pi by 2 is equal to 0.

00:29 So to solve this let us take t is equal to sorry a is equal to t minus pi by 2 and our equation will become y double dash of a plus pi by 2 plus 6y dash of a plus pi by 2 plus 8 into a plus pi by 2 which will be equal to minus 3 cos a plus pi by 2 and minus 2 sin we have a plus pi by 2.

01:09 So this can be again written as taking y of a plus pi by 2 which is equal to u of a and y dash of a plus pi by 2 can be taken as u dash of a and y double dash of a plus pi by 2 can be taken as u double dash of a so that the above equation can be reduced as u double dash of a plus 6 u dash of a plus 8 u of a equal to minus 3 cos a plus pi by 2 minus 2 sin a plus pi by 2.

02:07 So again we can take u of 0 is equal to 1 and u dash of 0 is equal to 0 according to the initial condition.

02:20 Now applying laplace on both sides we get the equation as l of u double dash of a plus 6 l of u dash of a plus 8 l of u of a which is equal to minus 3 l of cos a plus pi by 2 minus 2 l of sin a plus pi by 2.

03:08 Now so on applying laplace this can be again written as s squared u of s minus s u of 0 u of 0 minus u dash of 0 plus 6 s u of s minus u of 0 plus 8 u of s which will be equal to so here applying laplace on cos a plus pi by 2 this will become minus 3 into minus 1 by s squared plus 1 and this sine value will become minus 2 s by s squared plus 1...

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