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