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Q26E

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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 180
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

Find a particular solution to the differential equation.

y''+2y'+2y=4te-tcost

The particular solution is yp=e-t[tcost+t2sint].

See the step by step solution

Step by Step Solution

Step 1: Firstly, write the auxiliary equation of the above differential equation.

The differential equation is:

y''+2y'+2y=4te-tcost                         ......(1)

Write the homogeneous differential equation of the equation (1),

y''+2y'+2y=0

The auxiliary equation for the above equation,

m2+2m+2=0

Step 2: Now find the roots of the auxiliary equation.

Solve the auxiliary equation,

m2+2m+2=0m=-2±22-4(1)(2)2(1)m=-2±-42m=-1±i

The roots of the auxiliary equation are,

m1=-1+i,   &  m2=-1-i

The complementary solution of the given equation is,

yc=e-x(c1cosx+c2sinx)

Step 3: Use the method of undetermined coefficients to find a particular solution to the differential equation

Assume, the particular solution of equation (1),

yp=t[e-t(At+B)cost+e-t(Ct+D)sint]yp=e-t[At2cost+Btcost+Ct2sint+Dtsint]                  ......(2)

Now find the derivative of the above equation,

role="math" localid="1654875111330" yp'=-e-t[At2cost+Btcost+Ct2sint+Dtsint]         +e-t[2Atcost-At2sint+Bcost-Btsint+2Ctsint+Ct2cost+Dsint+Dtcost]yp'=e-t[(-A+C)t2cost+(-B+D+2A)tcost+(-C-A)t2sint+(-D-B+2C)tsint+Bcost+Dsint]yp''=-e-t(-A+C)t2cost+(-B+D+2A)tcost+(-C-A)t2sint+(-D-B+2C)tsint+Bcost+Dsint+e-t2(-A+C)tcost-(-A+C)t2sint+(-B+D+2A)cost-(-B+D+2A)tsint+2(-C-A)tsint+(-C-A)t2cost+(-D-B+2C)sint+(-D-B+2C)tcost-Bsint+Dcostyp''=e-t(-2C)t2cost+(-2D-4A+4C)tcost+(2A)t2sint+(2B-4C-4A)tsint+(-2B+2D+2A)cost+(-2D-2B+2C)sint

From the equation (1), Substitute the value of y''p, y'p and yp in the equation (1),

role="math" localid="1654875265020" yp''+2yp'+2yp=4te-tcoste-t(-2C)t2cost+(-2D-4A+4C)tcost+(2A)t2sint+(2B-4C-4A)tsint+(-2B+2D+2A)cost+(-2D-2B+2C)sint+2e-t[(-A+C)t2cost+(-B+D+2A)tcost+(-C-A)t2sint+(-D-B+2C)tsint+Bcost+Dsint]+2e-t[At2cost+Btcost+Ct2sint+Dtsint]=4te-tcoste-t[(4C)tcost-(4A)tsint+(2D+2A)cost+(-2B+2C)sint]=4te-tcost

Step 4: Final conclusion.

Comparing all coefficients of the above equation;

4C=4    C=14A=0    A=02D+2A=0                         .....(3)-2B+2C=0                      ......(4)

Substitute the value of A in the equation (3),

2D+2(0)=0D=0

Substitute the value of C in the equation (4),

-2B+2(1)=0B=1

Therefore, the particular solution of equation (1),

yp=e-t[At2cost+Btcost+Ct2sint+Dtsint]yp=e-t[tcost+t2sint]

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