Given the D.E y′′′+y′′=8x2
The auxiliary equation is of the form m3+m2=0
Equivalently we have m2(m+1)=0 .......(1)
Solving (1) we have m=−1 or m=0 (twice)
Then the complimentary function is of the form : Yc(x)=A1e−x+A2x+A3
Where A1,A2,A3 are constants.
Assuming the general form of the particular integral of the RHS of the DE:
Yp=Ax4+Bx3+cx2
Yp′=4Ax3+3Bx2+2cxYp′′=12Ax2+6Bx+2cYp′′′=24Ax+6BThenYp′′′+Yp′′=12Ax2+6Bx+24Ax+6B+2c=8x2
By the comparison of coefficients we have
12A=8; 6B+24A=0; 6B+2c=0
Solving the equations simultaneously we have:
A=32;B=−38;c=8
Therefore, Yp=Yp=32x4−38x3+8x2
The general solution of the DE is now of the form:
Y(x)=Yc(x)+Yp(x)=(A1e−x+A2x+A3)+32x4−38x3+8x2
Y(x)=A1e−x+32x4−38x3+8x2+A2x+A3
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