Question

 (3-y)/x^2  dx+(y^2- 2x)/(xy^2 ) dy=0        y(-1)=2

Answer 1

The given differential equation is $\frac{3-y}{x^2}dx \space +\frac{y^2-2x}{xy^2}dy=0$ .

The given differential equation is of first order and first degree and can be expressed in the form of

$Mdx \space + Ndy =0$ ,where M and N are functions of x and y .

compairing with standard form we get $M = \frac{3-y}{x^2} \space and \space N = \frac{y^2-2x}{xy^2}$

$Now \space \frac{\delta M}{\delta y }= -1/x^2$ ( partially differentiating M with respect to y keeping x as constant)

$and \space \frac{\delta N}{\delta x }= \frac {\delta }{\delta x}(y^2/xy^2 - 2x/xy^2)\\or, \frac{\delta N}{\delta x }= \frac {\delta }{\delta x}(1/x - 2/y^2)\\or, \frac{\delta N}{\delta x }=-1/x^2$( partially differentiating N with respect to x keeping y as constant)

$As \space \space \frac{\delta M}{\delta y }= \frac{\delta N}{\delta x }$

$\therefore$ The given differential equation is exact.

If the equation is exact ,then for some function $u(x,y)$ we have

$Mdx \space + Ndy =$ a perfect differential of $u = du.$

Now the general solution can be obtained by the following rules :

1. integrate M dx as if y is constant
2. integrate N dy as if x is constant
3. delete the term from 2 which is already obtained in 1.
4. the sum of these integrals will give a function $u(x,y)$ such that $du = M dx + N dy.$

1) $\int ( \frac{3-y}{x^2}) dx = (3-y) \int \frac{1}{x^2}dx$

$= (3-y)\frac{(-1)}{x}$

$= y/x \space - 3/x.$

2) $\int \frac{y^2-2x}{xy^2} dy = \int (y^2/xy^2 - 2x/xy^2) dy$

$=\int (1/x \space - 2/y^2) dy$

$= \int(1/x) dy \space - 2 \int(1/y^2)dy$

$= y/x \space + 2/y.$

3) $u = \frac{y}{x} - \frac{3}{x} + \frac{2}{y}$ (deleting the term (y/x) already obtained in 1.)

4) The required primitive is

$\frac{y}{x} - \frac{3}{x} + \frac{2}{y} = C$

$Now \space when \space x=-1, y = 2.$

Therefore $\frac{2}{-1} - \frac{3}{(-1)} + \frac{2}{2} = C$

$\therefore C = -2 +3 +1 =2$

The final solution is

$\frac{y}{x} - \frac{3}{x} + \frac{2}{y} = 2$ (ANSWER)