Question

y′ =y/x(x-x^3) ; y = −2 when x = 2

Answer 1

$\displaystyle y' = \frac{y}{x(x-x^3)} \\= \frac{y'}{y}= \frac{1}{x(x-x^3)} \\\text{Next, we integrate the expression above, to do this we resolve $\frac{1}{x(x-x^3)}$ using partial } \\\text{fractions} \\\implies \frac{1}{x(x-x^3)}=\frac{1}{2(1-x)}+\frac{1}{2(1+x)}+\frac{1}{x^2} \\\implies iny =\frac{1}{2}\int \frac{1}{1-x} + \frac{1}{2}\int \frac{1}{1+x} + \int \frac{1}{x^2} \\= iny = -\frac{1}{2}in(1-x)+\frac{1}{2}in(1+x)-x^{-1}+c \\\text{Taking the exponent of the expression above} \\y = (\frac{1+x}{1-x})^\frac{1}{2}-e^{x^{-1}}+A \\\text{Substituting y = -2, x= 2, we have that} \\ A=-0.351-1.73i \\\implies y = (\frac{1+x}{1-x})^\frac{1}{2}-e^{x^{-1}}-0.351-i.73i$