Solve differential equation 6(y-x^2)dx+xdy=0

$6(y-x^2)dx+xdy=0$
$(6(y-x^2)dx)/(xdx)+(xdy)/(xdx)=0$
$(6(y-x^2)dx)/(xdx)+(xdy)/(xdx)=0$
$6((y-x^2)/x)+dy/dx=0$ $dy/dx+(6/x)y=6x$ (1)
Now, equation (1) is first order linear differential equation
We know that solution of first order linear differential equation is given by
$y.(I.F.)=\int (I.F.)6xdx$ (2)x
where, $I.F.=e^{\int 6/xdx}$
$I.F.=e^{6\int dx/x}$ $({:}^{\prime }\int dx/x=\ln x+c)$
$=e^{6(\ln x)}$
$=e^{\ln x^6}$
$=x^6$
Now, solving equation(2) by using I.F.
$y(x^6)=\int x^6(6x)dx$
$=6\int x^{7}dx$ $({:}^{\prime }\int x^{n}dx=(x^{n+1})/(n+1)+c$
$=6(x^8/8)+c$
$=(3x^8)/4+c$ (3)
Now, using y(1)= 1 in equation (3)
$1(1{)}^6=(3(1{)}^8)/4+c$
$1=3/4+c$
$c=1-3/4$
$c=(4-3)/4$
$c=1/4$
Now, putting value of c in equation (3) $y{x}^6=(3x^8)/4+1/4$
Hence, solution of given differential equation is $y{x}^6=(3x^8)/4+1/4$