Question

y' + [1-3x^2/x(1-x^2)]y = 1/1-x^2

Answer 1

Given differential equation is $y'+ \frac{1-3x^2}{x(1-x^2)}y=\frac{1}{1-x^2} (\clubs)$ , clearly, it's a first order first and degree

differential equation of the form $y'+P(x)y=Q(x)$ ,Thus integrating factor $(I.F)$ is given by,

$$I.F=e^{\int P(x) dx}$$

Hence, In this case I.F will be,

$$I.F=e^{\int \frac{1-3x^2}{x(1-x^2)}dx} \hspace{1.5cm} (\star)$$

Now, let's calculate the integral first as follows

$$\int \frac{1-3x^2}{x(1-x^2)}dx \hspace{1cm} (\star \star)$$

Suppose, $f(x)=x(1-x^2)$ ,clearly $f'(x)=1-3x^3$ ,thus above integral has the form

$$\int \frac{f'(x)}{f(x)}dx$$

And we already know that $\int \frac{f'(x)}{f(x)}dx=ln(f(x)) +constant$

Hence, $(\star \star)$ will be $ln(x(1-x^2))$ ,thus from $(\star)$

$$I.F=e^{ln(x(1-x^2)} =x(1-x^2)$$

$\big(\because \: e^{ln(f(x))}=f(x)\big)$ ,Now on multiply I.F in equation $(\clubs)$ we get,

$$x(1-x^2)y'+(1-3x^2)y=\frac{x(1-x^2)}{1-x^2}\\ \implies \frac{d}{dx}(y \cdot x(1-x^2))=x\\ \implies d(y \cdot x(1-x^2))=x dx\\ \implies \int d(y \cdot x(1-x^2))=\int x dx\\ \implies y \cdot x(1-x^2)=\frac{x^2}{2} +c\\ \implies y=\frac{x}{2(1-x^2)}+\frac{c}{x(1-x^2)}$$

where $c$ is a constant. Hence, we are done.