Answer to Question #263196 in Differential Equations for Lotus

Question

Obtain the differential equation by eliminating arbitrary constant.

A ln y = Be^x²

Answer 1

$A ln y = Be^{x^2} ...(1)\\$

Differentiating both sides w.r.t $x$ , we get

$\frac{A}{y}.y'=2xBe^{x^2}\\ \frac{A}{y}.y'=2x.A ln y\ [\text{From eqn. (1)}]\\ y'=2xy.lny\\ y'-2xy.lny=0$