Answer to Question #191025 in Differential Equations for Rajanish Pathiwada

Question

1. ysin 2x dx-dy-y^2dy-cos^2xdy=0

Answer 1

The given differential equation is

$y sin2x dx -(1+y^2+cos^2x)dy=0$

Which is of the form $Mdx +Ndy =0$ .

Where 

$M=y sin2x \ \text{and} \ N=-(1+y^2+cos2x)$

$$\frac{\partial{M}}{\partial{y}}=\frac{\partial{ysin2x}}{\partial{y}}=sin2x$$ $$\frac{\partial{N}}{\partial{x}}=\frac{\partial{-(1+y^2+cos^2x)}}{\partial{x}}=-(2cosx×-sinx)$$

$=2sinxcos =sin2x$x

As $\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}}$ ,Hence the given differential equation is an exact differential equation.

$\therefore \$ The solution of the given differential equation is

$\int_{y=const.}Mdx+\int \text{(only those therms of N which do not contains x)}dy$

$=\int_{y=const.}ysin2x+\int{-(1+y^2)}dy$

$=y\int sin2x dx-\int(1+y^2)dy$

$=-\frac{ycos2x}{2}-y-\frac{y^3}{3}+c$

Where "c" is the integration constant