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

Question

$(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:

$\dfrac{d(x^2 - y^2 -1)}{2(x^2- y^2 -1)}= \dfrac{y dy}{(2x^2 + 3y^2 -7)}$ which is not useful at all.

Answer given is:

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

Answer 1

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$$\dfrac{dY}{dX}=\dfrac{2X+3Y-7}{3X+2Y-8}$$Again let us consider $$X=p+a$$$$Y=q+b$$$$\implies\dfrac{dq}{dp}=\dfrac{2p+2a+3q+3b-7}{3p+3a+2q+2b-8}=\dfrac{2p+3q}{3p+2q}$$From $2a+3b-7=0$ and $3a+2b-8=0$ we get $\implies a=2,b=1$ $$\dfrac{dq}{dp}=\dfrac{2p+3q}{3p+2q}\mbox{ is a homogeneous equation. So, the change of variable is given by}$$$$q(p)=p\ u(p)\implies u+s\dfrac{du}{dp}=\dfrac{2+3u}{3+2u}$$$$s\dfrac{du}{dp}=\dfrac{2+3u}{3+2u}-u=\dfrac{2-2u^2}{3+2u}$$$$\dfrac{dp}{p}=\dfrac{3+2u}{2-2u^2}du$$$$\ln|p|=\int\dfrac{3+2u}{2-2u^2}\ du=\dfrac14\ln|u+1|-\dfrac54-\ln|u-1|+C$$

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

Answer 2

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.

Answer 3

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.