Define the term homogeneous first order ordinary differential equation. Hence solve x^2 y' + xy = x2 + y

Rui Baldwin 2021-02-09 Answered
Define the term homogeneous first order ordinary differential equation.
Hence solve x2y+xy=x2+y
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Expert Answer

Maciej Morrow
Answered 2021-02-10 Author has 98 answers

dy/dx=F(y/x)
Now solve the given differential equation
x2y+xy=x2+y
y+(xy)/x2=(x2+y)/x2=>y+y/x=1+y/x2
y+y/xy/x2=1=>y+y(1/x1/x2)=1
P=1/x1/x2,Q=1
IF=ePdx=e(1/x1/x2)dx
=elogx+1/x=elogxe1/x
IF=xe1/x
Thus, the solution of differential equation is
y(IF)=IFQ(x)dx+C
Solving further to get
y(xe1/x)=xe1/x1dx+C
y(xe1/x)=xe1/xdx1e1/xdx+C
y(xe1/x)=(xe1/x)/(1/x2)+e1/x/(1/x2)dx+C
y(xe1/x)=x3e1/x+x2e1/xdx+C
=x3e1/x+x2e1/xdx+2xe1/xdx+C
=x3e1/x+1/3e1/xx3+1/6e1/xx2+1/6e1/xx1/6Ei(1/x)+C
Ei(x) is the exponential integral
y(xe1/x)=1/2e1/xx21/2(e1/xx+Ei(1/x))+C
y(x)=1/2x1/2(1+(Ei(1/x))/(xe1/x))+C/(xe1/x)

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Jeffrey Jordon
Answered 2021-10-23 Author has 2070 answers

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