Mathematics | VITEEE 2019 | Discussion Page

The solution of the differential equation $\left\{1+x\sqrt{(x^{2}+y^{2})}\right\}dx + \left\{\sqrt{(x^{2}+y^{2})}-1\right\}ydy = 0 is ?

$x^{2}+\frac{y^{2}}{2}+\frac{1}{3}(x^{2}+y^{2})^{3/2}$ =C

$x-\frac{y^{2}}{3}+\frac{1}{2}(x^{2}+y^{2})^{1/2}$ =C

$x-\frac{y^{2}}{2}+\frac{1}{3}(x^{2}+y^{2})^{3/2}$ =C

D) None of these

Option C

Rearranging the equation, we have

dx - ydy + $\sqrt{(x^{2}+y^{2})}(xdx+ydy) = 0$

$dx - ydy+\frac{1}{2}\sqrt{(x^{2}+y^{2})}d(x^{2}+y^{2}) = 0$

On integrating, we get

$x-\frac{y^{2}}{2}+\frac{1}{2}\int_{}^{}\sqrt{t}dt = C$

$\left\{t =\sqrt{x^{2}+y^{2}}\right\}$

or $x-\frac{y^{2}}{2}+\frac{1}{3}(x^{2}+y^{2})^{3/2}$ =C

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