Question

Find the solution of $\frac{2x^2+3y^2-7}{y}dx=\frac{3x^2+2y^2-8}{x}dy$.

• A. $x^2+y^2-3=\lambda (x^2+7^2-1)$
• B. $x^2+y^2-3=\lambda (3x^2+7^2-1)$
• C. $x^2+y^2-3=\lambda (3x^2-7^2-1)$
• D. $x^2+y^2-3=\lambda (x^2-7^2-1)$

Answer 1

The correct option is D $x^2+y^2-3=\lambda (x^2-7^2-1)$

$x^2=X,y^2=Y$
$\therefore 2xdx=dX$ and $2ydy=dY$
The given equation is $\frac{dY}{dX}=\frac{2X+3Y-7}{3X+2Y-8}$
Now proceed as usual $X=\alpha +h,Y=\beta +k$
$\frac{d\beta }{d\alpha }=\frac{2\alpha +3\beta }{3\alpha +2\beta }$ ...(1)
and $2h+3k-7=0$ and $3h+2k-8=0$
Solving, we have $h=2,k=1$
Now solve (1) as usual by putting $\beta =v\alpha$ etc.
$x^2+y^2-3=\lambda (x^2-7^2-1).$