Solve x^3y''' + x^2y'' - 2xy' + 2y = x^3lnx

Emeli Hagan 2020-12-09 Answered
Solve x3y+x2y2xy+2y=x3lnx
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Expert Answer

Mayme
Answered 2020-12-10 Author has 103 answers
Jeffrey Jordon
Answered 2021-09-30 Author has 2047 answers

Given: x3y+x2y2xy+2y=x3lnx

Let lnx=z then

dydx=dydzdzdx=1xdydx

Similarly:

x2d2ydx2=d2ydz2dydz and x3d3ydx3=d3ydz33d2ydz2+2dydx

Using above three expressions, the original differential equation reduces to:

(d3ydz33d2ydz2+2dydz)+(d2ydzdydz)2dydz+2y=ze3z

The auxilary equation of the differential equation

d3ydz32d2ydz2dydz+y=0 becomes

m32m2m+2=0

m=2,1,1

Therefore the complimentary function becomes:

yc(x)=ae2z+bez+cez

The potential integral is given by:

yp(x)=1D32D2D+2ze3z

=e3z1(D+3)32(D+3)2(D+3)+2z

=e3z1D3+7D2+14D+4z

=e3z14(1+D3+7D2+14D4)1z

=e3z4(1D3+7D2+14D4+...)z

=e3z4(z144)

=x34(lnx144)

Therefore the solution is

y(x)=yc(x)+yp(x)=ax2+bx+cx+x34(lnx144)

where a,b,c are arbitrary constants

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