Vtyjkyuk

$(2x^2 + 3y^2 -7)x dx = (3x^2 + 2y^2 -8 )y dy $

Attempt:

After expanding, everything is neat except: $3x^2y dy - 3y^2 x dx$

I can't convert it to exact differential.

Also, there's weird symmetry in the equation wrt the coefficients of $x^2$ and $y^2$. But not sure how to utilise that symmetry.

As an attempt, though, I reached this:

$dfracd(x^2 - y^2 -1)2(x^2- y^2 -1)= dfracy dy(2x^2 + 3y^2 -7)$ which is not useful at all.

Answer given is:

$(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C$

Given $(2x^2+3y^2-7)x dx=(3x^2+2y^2-8)y dy$

Let us take $$X=x^2$$$$Y=y^2$$and we get$$dfracdYdX=dfrac2X+3Y-73X+2Y-8$$Again let us consider $$X=p+a$$$$Y=q+b$$$$impliesdfracdqdp=dfrac2p+2a+3q+3b-73p+3a+2q+2b-8=dfrac2p+3q3p+2q$$From $2a+3b-7=0$ and $3a+2b-8=0$ we get $implies a=2,b=1$
$$dfracdqdp=dfrac2p+3q3p+2qmbox is a homogeneous equation. So, the change of variable is given by$$$$q(p)=p u(p)implies u+sdfracdudp=dfrac2+3u3+2u$$$$sdfracdudp=dfrac2+3u3+2u-u=dfrac2-2u^23+2u$$$$dfracdpp=dfrac3+2u2-2u^2du$$$$ln|p|=intdfrac3+2u2-2u^2 du=dfrac14ln|u+1|-dfrac54-ln|u-1|+C$$

$$s=Cdfrac(u+1)^frac14(u-1)^frac54implies(q-p)^frac54=C(q+p)^frac14$$
$$(q-p)^5=C(q+p)implies(q-p)^5-C(q+p)=0$$$$p=X-2mbox mbox and q=Y-1$$$$implies(Y-X+1)^5-C(Y+X-3)=0$$Therefore,$$boxed(x^2 +y^2 - 3)=(x^2 + y^2 - 1)^5C$$

Hint:
$$(2x^2 -4 + 3y^2 -3)x dx = (3x^2 -6+ 2y^2 -2 )y dy$$
and use substitutions $x^2-2=u$ and $y^2-1=v$. You will find a homogeneous differential equation.

The method to solve such differential equations is:

Substitute $x^2 = u implies 2x dx = du$ and $y^2 = v implies 2y dy = dv$

Then it's neat, a dy/dx = linear/linear type equation is obtained which can be solved using standard techniques of $y= Y+k$ and $x = X+h$ to eliminate the constant terms from both the linears.