Question

Solve:

$({3x}^2+y^2)dy+(x^2+{3y}^2)dx=0$

Answer 1

$(3x^2+y^2)dy+(x^2+3y^2)dx=0$

$\frac{dy}{dx}=-\frac{(x^2+3y^2)}{(3x^2+y^2)}$

$y=vx$

$\frac{dy}{dx}=v+x\frac{dv}{dx}$

$v+x\frac{dv}{dx}=-\frac{(x^2+3v^2{x}^2)}{(3x^2+v^2{x}^2)}$

$=-\left(\frac{1+3v^2}{3+v^2}\right)$

$x\frac{dv}{dx}=-\frac{(1+3v^2)}{(3+v^2)}-v$

$x\frac{dv}{dx}=-\frac{1-3v^2-3v-3v^3}{3+v^2}$

$x\frac{dv}{dx}=-\frac{(1+v{)}^3}{(3+v^2)}$

$\int \frac{3+v^2}{(1+v{)}^3}dx=\int \frac{dx}{x}$

$\int \frac{3+v^2}{(1+v{)}^3}dv+\int \frac{v^2}{(1+v{)}^3}dv=\int \frac{dx}{x}$

$3.\frac{(1+v{)}^{(-3+1)}}{(-3+1)}\frac{v^3}{(1+v{)}^3}-\int \frac{v^2[-v+2]}{(1+v{)}^4}dv\int -\frac{v^3}{(1+v{)}^4}dv+\int \frac{2v^2}{(1_v{)}^4}dv$

$\Rightarrow ln|x+1|+\frac{3}{x+1}-\frac{3}{2(x+1{)}^2}+\frac{1}{3(x+1{)}^3}+2[\frac{-x^2}{3(1+x{)}^3}+\frac{-2x-1}{3(x+1{)}^2}]$

$\Rightarrow -ln|x+1|-\frac{3}{x+1}+\frac{3}{2(x+1{)}^2}+\frac{3}{2(x+1{)}^2}-\frac{1}{3(x+1{)}^3}-\frac{2x^2}{3(1+x{)}^3}+\frac{2(-2x-1)}{3(x+1{)}^2}$

$\frac{x^3}{(1+x{)}^3}+ln(x+1)+\frac{3}{x+1}-\frac{3}{2(x+1{)}^2}+\frac{1}{3(x+1{)}^3}+\frac{2x^2}{3(1+x{)}^3}-\frac{2(-2x+1)}{3(x+1{)}^2}$

Replace x by (v) in all above equation By mistake written x, its (v) ,

So, we get : { and replace v by $\frac{y}{x}$ }

$\left\{\begin{array}{c}{\left(\frac{y}{x}\right)}^3\frac{1}{(1+y/x{)}^3}+ln(\frac{y}{x}+1)+\frac{3}{(y/x+1)}-\frac{3}{2(y/x+1{)}^2}\\ +\frac{1}{3(y/x+1{)}^3}+\frac{2\left(y/x\right)}{3(1+y/x{)}^3}+\frac{2\left(2\right(y/x)-1)}{3(y/x+1{)}^2}\\ =\log x+\log c\end{array}\right\}$