Question

determine the general/particular solution for each equation using the applicable solution to equations of order one (separable, homogenous,linear,exact)

1. (1+y^2)dx + (1=x^2)dy = 0 ; when x = 0, y = 1

Answer 1

Let us determine the general/particular solution of the differential equation

$(1+y^2)dx + (1+x^2)dy = 0 ;$ when $x = 0, y = 1.$

After dividing both parts by $(1+y^2) (1+x^2)$ we get the equation

$\frac{dx}{1+x^2} + \frac{dy}{1+y^2} = 0,$

and hence

$\int\frac{dx}{1+x^2} + \int\frac{dy}{1+y^2} = C.$

It follows that the general solution is of the form:

$\arctan x+\arctan y=C.$

Since when $x = 0, y = 1,$ we get $\arctan 0+\arctan 1=C,$ and hence $C=\frac{\pi}4.$

We conclude that the particular solution is the following:

$\arctan x+\arctan y=\frac{\pi}4.$