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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 350
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

In Problems 1-14, solve the given initial value problem using the method of Laplace transforms

y''-7y'+10y=9cost+7sint; y0=5, y'0=-4

The initial value for y''-7y'+10y=9cost+7sint is y(t)=-4e5t+8e2t+cost

See the step by step solution

Step by Step Solution

Step 1: Define the Laplace Transform

  • The Laplace transform is a strong integral transform used in mathematics to convert a function from the time domain to the s-domain.
  • In some circumstances, the Laplace transform can be utilized to solve linear differential equations with given beginning conditions.
  • Fs=0f(t)e-stt'

Step 2: Determine the initial value of Laplace transform

Applying the Laplace transform and using its linearity we get

Ly''-7y'+10y=L9cost+7sintLy''-7Ly'+10Ly=9ss2+1+7s2+1

Solve for the transform as:

s2Ys-sy0-y'0-7sYs-y0+10Ys=9s+7s2+1s2Ys-5s+4-7sYs-5+10Ys=s2-7s+10Ys=5s3-39s2+14s-32s2+1Ys=5s3-39s2+14s-32s2-7s+10s2+1

Solve for the partial fraction as:

5s3-39s2+14s-32s2-7s+10s2+1=5s3-39s2+14s-32s-5s-2s2+1=As-5+Bs-2+Cs+Ds2+1

Solve further as:

role="math" localid="1664299653224" 5s3-39s2+14s-32=As2+1s-2+Bs2+1s-5+Cs+Ds-5s-2

Using s=5,2,0,1 ,respectively, gives

s=5:-312=78AA=-4s=2:-120=-15BB=8s=0:-32=-32+10DD=0s=1:-52=4C-56C=1

Therefore, the equation is:

Ys=-4s-5+8s-2+ss2+1

Using the inverse Laplace transform we obtain the solution of given differential equation.

yt=L-1-4s-5+8s-2+ss2+1t=-4L1s-5+8L1s-2+Css2+1=-4e5t+8e2t+cost

Therefore, the solution of the initial value problem is:

y(t)=-4e5t+8e2t+cost

Therefore, the initial value for y''-7y'+10y=9cost+7sint is y(t)=-4e5t+8e2t+cost

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