Question

obtain the general and particular solution for differential equation below

(y - √x²+y²) dx - xdy = 0 when x=√3, y=1

Answer 1

Rearranging gives:dy

$\frac{dy}{dx}= \frac{y+ \sqrt{x^2+y^2}}{x}= \frac{y}{x}+ \sqrt{1+ (\frac{y}{x})^2}$

Try substituting u=y/x.

Then y′=u+xu′, giving:

$xu'= \sqrt{1+u^2}$

$\frac{u'}{ \sqrt{1+u^2}}= \frac{1}{x}$

sinh^-1u=c+ log(x)

u=sinh(c+ log(x))

General solution

$y=x sinh(c+ log(x))=x \frac{e^{c+log(x)}. - e^{-c-log(x)}}{2}=x \frac{e^ce^{log(x)}- 1/e^ce^{log(x)}}{2}$

$2= \sqrt{3}( \sqrt{3}e^c -1/ \sqrt{3}e^c) =3e^c -1/e^c$

$3e^{2c}-2e^c-1=0$

D=4+12=16

$e^c= \frac{2^+_- \sqrt{16}}{6}$

e^c₁=1

e^c₂= -1/3

Particular solution

$y_1=x \frac{e^{log(x)}-1/e^{log(x)}}{2}$

$y_2=x \frac{-1/3e^{log(x)}+ 3/e^{log(x)}}{2}$