Answer to Question #266809 in Differential Equations for Harsh

Question

Solve (2𝑥 + 𝑡𝑎𝑛𝑦)𝑑𝑥 + (𝑥 − 𝑥

2

𝑡𝑎𝑛𝑦)𝑑𝑦 = 0

Answer 1

Solution;

$(2x+tany)dx+(x-x^2tany)dy=0...(1)$

The equation is in the form;

$Mdx+Ndy=0$ ....(2)

Comparing (1) and (2);

$M=2x+tany$

$N=x-x^2tany$

$\frac{\partial M}{\partial y}=sec^2y$

$\frac{\partial N}{\partial x}=1-2xtany$

Therefore;

Using cos(y) as the integrating factor makes (1) exact.

Multiple (1) with the I.F;

$(2xcosy+siny)dx+(xcosy-x^2siny)dy=0.....(3)$A solution of (3) will be;

$\int_{y=constant}dx+\int$ (Terms in N without x)dy=C

$\int(2xcosy+siny)dx+\int0=C$

$cosy\int2xdx+siny\int1dx=C$

$x^2cosy+xsiny=C$