A nonhomogeneous second-order linear equation and a complementary

Kendall Daniels 2022-03-28 Answered
A nonhomogeneous second-order linear equation and a complementary function yc are given. Find a particular solution of the equation.
x2y4xy+6y=x3; yc=c1x2+c2x3
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Answers (1)

Gia Edwards
Answered 2022-03-29 Author has 12 answers
Calculation:
It is known that,
y1(x)=x2 and y2(x)=x3
Now, computing the Wro
ian,
W(x)=[(x2,x3),(ddx(x2),ddx(x3)]=[x2x32x3x2]=x4
Dividing the differential equation by the leading terms coefficient
x2y4xy+6y=x3
x2y4xy+6yx2=x3x2
y4yx+6yx2=x
Then,
f(x)=x
v1(x)=xy2(x)W(x)dx and v2(x)=xy2(x)W(x)dx
The particular solution is given by,
yp(x)=v1(x)y1(x)+v2(x)y2(x)
v1(x)=x(x3)x4dx=1dx=x
v2(x)=x(x2)x4dx=1xdx=lnx
yp(x)=x(x2)+(lnx)(x3)
yp(x)=x3(1+lnx)
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