Solve the linear differential equation of y(x^2+y)dx+x(x^2-2y)dy+0; when x=1, y=2

Linear Equation of Order One
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15 Answers

The given can be written as Pdx +Qdy=0
by Level 12 User (101k points)
It is not exact because Py-Qx=4y-2x^2
by Level 12 User (101k points)
→I•F=1/x^2
by Level 12 User (101k points)
Then p/x^2 dx+Q/x2 dy=0 is exact
by Level 12 User (101k points)
And du_= P/x^2 dx+Q/x^2 dy=0
by Level 12 User (101k points)
→ u(x,y)=c
by Level 12 User (101k points)
æu/æy=x-2y/x
by Level 12 User (101k points)
→u=w(x)+xy-y^2/x →(1)
by Level 12 User (101k points)
æu/æx=w'(x)+y+y^2/x^2_=y+y^2/x^2
by Level 12 User (101k points)
→w(x)=x →(1)
by Level 12 User (101k points)
(1),(2)→y^x-y^2/x+x=c with y(1)=2
by Level 12 User (101k points)
Therefore y^2-x^2•y-x(x+1)=0
by Level 12 User (101k points)
Solving in respect of Y with Determinant
by Level 12 User (101k points)
D=x^4+4x•(x+1)=(x^2+2)^2
by Level 12 User (101k points)
The solutions are: y={-1upper, x^2+1 lower
by Level 12 User (101k points)

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