Solution of Exam on Ordinary Differential Equations. Trial Aleph ...

3.1 <strong>Solution</strong>
2xy+y 2
x
= 2xy
2 x
+ y2
2 x
= 2 y 2 x +
y 2
x
, y
0
= 2 y
x +
y 2
x
;so equation (20) is
homogeneous equation.
y
x = z , y = zx; y0 = z 0 x + z; z 0 x + z = 2z + z 2 ; z 0 x = z + z 2 ;
dz
z+z
= dx; R dz
2 z+z 2
1
z+z
= 1 2 z
ln
z
(z+1)x
z = y x )
= R dx
x ;
1
z+1 )R dz
z+z
= R 1
2 z
= C;
z
(z+1)x
z
z+1 = y x
y
x +1 =
dz
dx x = z + z2 (21)
1
z+1
dz = ln jzj
ln jz + 1j = ln
z
z+1
ln
z
z + 1 = ln jxj + C (22)
= e c = jCj
y
x+y ;
z
(z + 1) x = C
z
(z+1)x =
y
(x+y)x ;
y
(x + y) x = C
y = Cx (x + y) = Cxy + Cx 2 ; y Cxy = Cx 2 ; y (1 Cx) = Cx 2
y (1) = 1 : 1 y
2
= ln 1 + C = C;
y =
Cx2
1 Cx
(x+y)x = 1 2 () y =
x2 = 2x2
1 1 2 x 2 x
3.2 Answer
y
(x + y) x = 1 2x2
() y =
2 2 x
(23)
4 Problem 3-bet
4.1 <strong>Solution</strong>.
Equation (24) is an equation <strong>of</strong> the form
where p (x) = 1; q (x) = e x ; m = 3:
y (x) = 0 is a solution (partial or singular?)
Now assume y 6= 0:
y 0 + y = e x y 3 (24)
y 0 + p (x) y = q (x) y m ; (25)
4