Question

The general solution of the differential equation (x – y2)dx + y(5x + y2)dy = 0 is:

• A. (y2+x)4=C|(y2+2x)3|
• B. (y2+2x)4=C|(y2+x)3|
• C. |(y2+x)3|=C(2y2+x)4
• D. |(y2+2x)3|=C(2y2+x)4

Answer 1

Answer: (A)

(y2+x)4=C|(y2+2x)3|

Answer 2

(x−y2)dx+y(5x+y2)dy=0
y $\frac{dy}{dx}$=y2−$\frac{x}{5}$x+y2
Let y2 = t
$\frac{1}{2}$⋅$\frac{dt}{dx}$=t−$\frac{x}{5}$x+t
Now substitute, t = vx
$\frac{dt}{dx}$=v+x $\frac{dv}{dx}$
$\frac{1}{2}${v+x $\frac{dv}{dx}$}=$\frac{v-1}{5+v}$
x $\frac{dv}{dx}$=$\frac{2v-2}{5+v}-v$=$\frac{3v-v^2-2}{5+v}$
$\int \frac{4}{v+1}dv-\int\frac{3}{v+2}dv=-\int\frac{dx}{x}$
$|\frac{(v+1)^4}{(v+2)^3}|=\frac{C}{x}$
$\frac{(\frac{y^2}{x}+1)^4}{(\frac{y^2}{x}+2)^3}$=$\frac{C}{x}$
$(y^2+x)^4$=$C|y^2+2x|^3$