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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 186
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 general solution to the differential equation.y''(x)+6y'(x)+10y(x)=10x4+24x3+2x2-12x+18

The general solution to the given differential equation is:

y=c1e-3xcosx+c2e-3xsinx+x4-x2+2.

See the step by step solution

Step by Step Solution

Step 1: Write the auxiliary equation of the given differential equation. 

The differential equation is,

y''(x)+6y'(x)+10y(x)=10x4+24x3+2x2-12x+18                      (1)

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

y''(x)+6y'(x)+10y(x)=0

The auxiliary equation for the above equation,

m2+6m+10=0

Step 2: Find the complementary solution of the given equation.

Solve the auxiliary equation,

m2+6m+10=0m=-6±36-402m=-6±42m=-3±i

The roots of the auxiliary equation are,

m1=-3+i,      m2=-3-i

The complementary solution of the given equation is,

yc=c1e-3xcosx+c2e-3xsinx

Step 3: Now find the particular solution to find a general solution for the equation.

Assume, the particular solution of equation (1),

yp(t)=Ax4+Bx3+Cx2+Dx+E                  ......(2)

Now find the first and second derivatives of the above equation,

yp'(t)=4Ax3+3Bx2+2Cx+Dyp''(t)=12Ax2+6Bx+2C

Substitute the value of yp(t),  yp'(t) and yp''(t) the equation (1),

y''(x)+6y'(x)+10y(x)=10x4+24x3+2x2-12x+1812Ax2+6Bx+2C+6(4Ax3+3Bx2+2Cx+D)+10(Ax4+Bx3+Cx2+Dx+E)       =10x4+24x3+2x2-12x+1810Ax4+(10B+24A)x3+(12A+18B+10C)x2+(6B+12C+10D)x+(2C+6D+10E)       =10x4+24x3+2x2-12x+18

Comparing all coefficients of the above equation,

10A=10    A=110B+24A=24                          (3)12A+18B+10C=2                             (4)6B+12C+10D=-12                        (5)2C+6D+10E=18                         (6)

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

10B+24(1)=24B=0

Substitute the value of A and B in the equation (4),

12(1)+18(0)+10C=2C=-1

Substitute the value of C and B in the equation (5),

6(0)+12(-1)+10D=-12D=0

Substitute the value of C and D in the equation (6),

2(-1)+6(0)+10E=18E=2

Substitute the value of A, B, C, D, and E in the equation (2),

yp(t)=Ax4+Bx3+Cx2+Dx+Eyp(t)=(1)x4+(0)x3+(-1)x2+(0)x+(2)yp(t)=x4-x2+2

Therefore, the particular solution of equation (1),

yp(t)=x4-x2+2

Step 4: Conclusion

Therefore, the general solution is,

y=yc(t)+yp(t)y=c1e-3xcosx+c2e-3xsinx+x4-x2+2

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