Question

`y'= (1)/(6) x(1- y^(2) )`

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

`(\mathrm dy)/(\mathrm dx)=\frac16x(1-y^2)`

`\displaystyle\int(\mathrm dy)/(1-y^2)=\frac16\int x\,\mathrm dx`

`\frac12\ln|1+y|-\frac12\ln|1-y|=\frac1{12}x^2+C`

`\ln\left|(1+y)/(1-y)\right|=\frac16x^2+C`

`(1+y)/(1-y)=e^(x^2/6+C)`

`-1+\frac2{1-y}=Ce^(x^2/6)`

`y=1-\frac2{1+Ce^(x^2/6)}`

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


`7=1-\frac2{1+Ce^0}\implies C=-\frac43`

so the particular solution is


`y=1-\frac2{1-\frac43e^(x^2/6)}`