Question

Laplace transform (D²-3D+2)y=4 given y(0)=2 and y’(0)=3​

Answer 1

SOLUTION

TO SOLVE

Using Laplace transform

(D²-3D+2)y=4 given y(0)=2 and y'(0)=3

EVALUATION

Here the given equation is

(D²-3D+2)y=4

Taking Laplace transform in both sides we get

$\displaystyle \sf{[ {s}^{2} \bar{y} - sy(0) - y'(0) ] - 3[ s \bar{y} - y(0) ] + 2\bar{y} = \frac{4}{s} }$

$\displaystyle \sf{ \implies \: [ {s}^{2} \bar{y} - 2s - 3 ] - 3[ s \bar{y} - 2 ] + 2\bar{y} = \frac{4}{s} }$

$\displaystyle \sf{ \implies \: ({s}^{2} - 3s + 2)\bar{y} - 2s - 3 + 6= \frac{4}{s} }$

$\displaystyle \sf{ \implies \: ({s}^{2} - 3s + 2)\bar{y} = \frac{4}{s} + 2s - 3}$

$\displaystyle \sf{ \implies \: ({s}^{2} - 3s + 2)\bar{y} = \frac{4 + 2 {s}^{2} - 3s }{s} }$

$\displaystyle \sf{ \implies \: ({s}^{2} - 3s + 2)\bar{y} = \frac{2( {s}^{2} - 3s + 2) + 3s }{s} }$

$\displaystyle \sf{ \implies \: \bar{y} = \frac{2( {s}^{2} - 3s + 2) + 3s }{s({s}^{2} - 3s + 2)} }$

$\displaystyle \sf{ \implies \: \bar{y} = \frac{2}{s} + \frac{3 }{({s}^{2} - 3s + 2)} }$

$\displaystyle \sf{ \implies \: \bar{y} = \frac{2}{s} + \frac{3 }{(s - 1)(s - 2)} }$

$\displaystyle \sf{ \implies \: \bar{y} = \frac{2}{s} + \frac{3[(s - 1) - (s - 2)] }{(s - 1)(s - 2)} }$

$\displaystyle \sf{ \implies \: \bar{y} = \frac{2}{s} + \frac{3}{s - 2} - \frac{3}{s - 1} }$

On inversion we get

$\displaystyle \sf{ \implies \: y = {L}^{ - 1} \bigg( \frac{2}{s} \bigg) + {L}^{ - 1} \bigg(\frac{3}{s - 2}\bigg) - {L}^{ - 1} \bigg(\frac{3}{s - 1}\bigg) }$

$\displaystyle \sf{ \implies \: y = 2{L}^{ - 1} \bigg( \frac{1}{s} \bigg) + 3 {L}^{ - 1} \bigg(\frac{1}{s - 2}\bigg) - 3 {L}^{ - 1} \bigg(\frac{1}{s - 1}\bigg) }$

$\displaystyle \sf{ \implies \: y = 2 + 3 {e}^{2t} - 3 {e}^{t} }$

Which is the required solution

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