Question

Find the general/particular solution of the following Differential Equations.

(Exact D.E)

2.) (2xy-tany)dx+(x²-xSec²y)dy=0

Answer 1

Solution;

Check for exactness of the equation;

$M=2xy-tany$

$N=x^2-xsec^2y$

$\frac{dM}{dy}=$ $2x-sec^2y$

$\frac{dN}{dx}=2x-sec^2y$

Clearly;

$\frac{dM}{dy}=\frac{dN}{dx}$

The equation is exact.

Solution is given as;

$\int_{y=c}Mdx+\int$ (Terms of N independent of x)$dy$ $=C$

$\int(2xy-tan y)dx+0=C$

$2y\int xdx-tan(y)\int 1dx=C$

$2y×(\frac{x^2}{2})-tany(x)=C$

The general solution is;

$x^2y-xtan(y)=C$