How to solve this equation y‴−8y"+4y′+48=4x2+1 using Laplace Transformation.

Step 1 Converting the equation in Laplace transform: s3Y−8s2Y+4sY+48Y=4d2ds21s+1s s3Y−8s2Y+4sY+48Y=8s3+1s s3Y−8s2Y+4sY+48Y=(8+s2)(s3) (s3−8s2+4s+48)Y=(8+s2)(s3) Y=(8+s2)s3(s3−8s2+4s+48) Step 2 Converting into partial fraction and solving: Y=(8+s2)s3(s3−8s2+4s+48) =(8+s2)s3(s−4)(s+2)(s−6) =1(6s3)−1(72s2)−132(s−4)+43864s−132(x+2)+11864(s−6) =112d2ds21s+−172dds1s−132(s−4)+43864s−132(s+2)+11864(s−6) Step 3 Thus finding inverse Laplace transform find y(t): y(t)=(x2)12+x72−e4x32+43864−e−2x32+11e6x864

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CMIIh

Answered question

2020-11-23

How to solve this equation

y^{‴} - 8 y " + 4 y^{'} + 48 = 4 x^{2} + 1

using Laplace Transformation.

Answer & Explanation

Macsen Nixon

Skilled2020-11-24Added 117 answers

Step 1
Converting the equation in Laplace transform:

s^{3} Y - 8 s^{2} Y + 4 s Y + 48 Y = 4 \frac{d^{2}}{d s^{2}} \frac{1}{s} + \frac{1}{s}

s^{3} Y - 8 s^{2} Y + 4 s Y + 48 Y = \frac{8}{s^{3}} + \frac{1}{s}

s^{3} Y - 8 s^{2} Y + 4 s Y + 48 Y = \frac{(8 + s^{2})}{(s^{3})}

(s^{3} - 8 s^{2} + 4 s + 48) Y = \frac{(8 + s^{2})}{(s^{3})}

Y = \frac{(8 + s^{2})}{s^{3} (s^{3} - 8 s^{2} + 4 s + 48)}

Step 2
Converting into partial fraction and solving:

Y = \frac{(8 + s^{2})}{s^{3} (s^{3} - 8 s^{2} + 4 s + 48)}

= \frac{(8 + s^{2})}{s^{3} (s - 4) (s + 2) (s - 6)}

= \frac{1}{(6 s^{3})} - \frac{1}{(72 s^{2})} - \frac{1}{32 (s - 4)} + \frac{43}{864 s} - \frac{1}{32 (x + 2)} + \frac{11}{864 (s - 6)}

= \frac{1}{12} \frac{d^{2}}{d s^{2}} \frac{1}{s} + \frac{- 1}{72} \frac{d}{d s} \frac{1}{s} - \frac{1}{32 (s - 4)} + \frac{43}{864 s} - \frac{1}{32 (s + 2)} + \frac{11}{864 (s - 6)}

Step 3
Thus finding inverse Laplace transform find y(t):

y (t) = \frac{(x^{2})}{12} + \frac{x}{72} - \frac{e^{4 x}}{32} + \frac{43}{864} - \frac{e^{- 2 x}}{32} + \frac{11 e^{6 x}}{864}

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