Question

1. Find the general solution of the differential equation dy/dx = y^2 /1 + x^2

Answer 1

The given question is; $\displaystyle \frac{dy}{dx}=\frac{y^2}{1+x^2}$

Now, by method of separation of variables we have;

$\displaystyle \frac{1}{y^2}\frac{dy}{dx}=\frac{1}{1+x^2}$

integrating both sides with respect to x yields;

$\displaystyle \int\displaystyle \frac{1}{y^2}\frac{dy}{dx}\ dx=\int\frac{1}{1+x^2}\ dx\\ \Rightarrow\int y^{-2}\ dy=\int\frac{1}{1+x^2}\ dx\\ \text{But}\int\frac{1}{1+x^2}\ dx=\tan^{-1}(x)\ \text{is standard.}\\ \Rightarrow \frac{y^{-2+1}}{-2+1}=\tan^{-1}(x)+c,\ \text{where c is an arbitrary constant.}\\ \Rightarrow\frac{y^{-1}}{-1}=\tan^{-1}(x)+c\\ \Rightarrow -\frac{1}{y}=\tan^{-1}(x)+c\\ \Rightarrow y=\frac{-1}{\tan^{-1}(x)+c\\}$

Hence, the general solution of the given DE is;

$\displaystyle y=\frac{-1}{\tan^{-1}(x)+c\\}$