Answer to Question #140898 in Differential Equations for Victor

Question #140898

y'''+y''=8x^2


1
Expert's answer
2020-10-28T19:35:14-0400

Given the D.E y+y=8x2y'''+y''=8x^2

The auxiliary equation is of the form m3+m2=0m^3+m^2=0

Equivalently we have m2(m+1)=0m^2(m+1)=0 .......(1)

Solving (1) we have m=1m=-1 or m=0m=0 (twice)

Then the complimentary function is of the form : Yc(x)=A1ex+A2x+A3Y_c(x)=A_1e^{-x}+A_2x+A_3

Where A1,A2,A3A_1,A_2,A_3 are constants.

Assuming the general form of the particular integral of the RHS of the DE:

Yp=Ax4+Bx3+cx2Y_p = Ax^4+Bx^3+cx^2

Yp=4Ax3+3Bx2+2cxYp=12Ax2+6Bx+2cYp=24Ax+6BThenYp+Yp=12Ax2+6Bx+24Ax+6B+2c=8x2Y'_{p} = 4Ax^3+3Bx^2+2cx\\ Y''_{p} = 12Ax^2+6Bx+2c\\ Y'''_{p} = 24Ax+6B\\ \\Then\\ Y'''_p+Y''_p=12Ax^2+6Bx+24Ax+6B+2c=8x^2


By the comparison of coefficients we have

12A=8;12A=8 ; 6B+24A=0;6B+24A=0; 6B+2c=06B+2c= 0

Solving the equations simultaneously we have:

A=23;B=83;c=8A=\frac{2}{3};B=-\frac{8}{3};c=8

Therefore, Yp=Yp=23x483x3+8x2Y_p=Y_p = \frac{2}{3}x^4-\frac{8}{3}x^3+8x^2

The general solution of the DE is now of the form:

Y(x)=Yc(x)+Yp(x)=(A1ex+A2x+A3)+23x483x3+8x2Y(x)=Y_c(x)+Y_p(x)= (A_1e^{-x}+A_2x+A_3)+\frac{2}{3}x^4-\frac{8}{3}x^3+8x^2

Y(x)=A1ex+23x483x3+8x2+A2x+A3Y(x)=A_1e^{-x}+\frac{2}{3}x^4-\frac{8}{3}x^3+8x^2+A_2x+A_3


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