Question

c{1}{6} x(1- y^{2} )" alt="y'= \frac{1}{6} x(1- y^{2} )" align="absmiddle" class="latex-formula">
This equation has the 2 constant solutions (in increasing order), y=? and y=?
The solution of this equation subject to the initial condition y(0)=7y(0)=7 is ?

Answer 1

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac16x(1-y^2)$
$\displaystyle\int\frac{\mathrm dy}{1-y^2}=\frac16\int x\,\mathrm dx$
$\dfrac12\ln|1+y|-\dfrac12\ln|1-y|=\dfrac1{12}x^2+C$
$\ln\left|\dfrac{1+y}{1-y}\right|=\dfrac16x^2+C$
$\dfrac{1+y}{1-y}=e^{x^2/6+C}$
$-1+\dfrac2{1-y}=Ce^{x^2/6}$
$y=1-\dfrac2{1+Ce^{x^2/6}}$

Given that $y(0)=7$, we get

$7=1-\dfrac2{1+Ce^0}\implies C=-\dfrac43$

so the particular solution is

$y=1-\dfrac2{1-\frac43e^{x^2/6}}$