Question

y(x^2 + y^2)dx + x(3x^2 - 5y^2)dy=0 when x=1, y=1

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Answer 1

Let y=vx, then dy/dx=v+xdv/dx.

The DE becomes:

vx(x²+v²x²)+x(3x²-5v²x²)(v+xdv/dx)=0,

v(1+v²)+(3-5v²)(v+xdv/dx)=0,

v+v³+3v+3xdv/dx-5v³-5v²xdv/dx=0,

4v-4v³=-x(3-5v²)dv/dx,

∫dx/x=¼∫[(5v²-3)/(v-v³)]dv.

The variables are now separated so integration should be easier.

The right hand integral can be split into partial fractions: -3/v+2v/(1-v²) which is easily integrated. So:

ln|x|=-3ln|v|-ln|1-v²|+A where A is a constant.

ln|x|=-ln|v³(1-v²)|+A. If A=ln(C) where C is a constant then:

ln|x|=ln(C/(v³-v⁴)) and x=C/(v³-v⁴).

But v=y/x, so x=C/((y/x)³-(y/x))=Cx³/(y³-x²y), Cx²=y³-x²y.

Initial conditions are (x,y)=(1,1) so C=0, y²=x².

Now check out the original DE using dy/dx=x/y.

y²(x²+y²)+x²(3x²-5y²)=0, 2x⁴-2x⁴=0, which is true. So this confirms the solution y²=x².