Question

solve the following DE by using the appropriate substitution:

(y^2 + yx) dx - x^2 dy = 0

Answer 1

( xy + y^2 ) dx + x^2 dy = 0

(xy + y^2 ) dx = - x^2 dy

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

On RHS, separating,

dy/dx = - ( y/x) - (y/x)^2............(1)
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Substitute y/x = v so that y = vx.

Then, dy/dx = v + x ( dv/dx).
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Hence, eq(1) now becomes

v + x( dv/dx ) = - v - v^2

x( dv/dx ) = - 2v - v^2 = - v( 2 + v )

dv / v( 2 + v ) = - 2 dx / x

Integrating both sides

( 1/2) INT [ 1/v - 1/ ( 2 + v ) ] dv = - 2 INT ( 1 / x ) dx

ln v - ln ( 2 + v ) = - 4 ln x + 2 ln c.....Note the constant....

ln [ v / ( 1 + v ) ] = ln [ c^2 / x^2 ]

v / ( 2 + v ) = c^2 / x^4

(y/x) / ( 2 + y/x ) = C / x^4

y / (2 x + y ) = C / x^4

x^4. y = C (2x + y ). ................General Solution.

Answer 2

( y² + xy ) dx = x² dy

∴ dy/dx = ( y² + xy ) / x²

∴ dy/dx = (y/x)² + (y/x) ... (1)
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Put ... y/x = v, ... i.e., ... y = vx.

∴ dy/dx = v + x(dv/dx).
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From (1), then,

... v + x(dv/dx) = v² + v

∴ x (dv/dx) = v²

∫ (1/v² ) dv = ∫ (1/x) dx

∴ - 1/v = ln |x| + C

∴ - x/y = ln |x| + C .......... Ans.

Answer 3

don't you have to prove that it is homegeneous at first by proving M(tx,ty) = t^n M(x,y)?

Answer 4

i think you do ? thats a good question