Question

Mohon bantuan kk ......................

Answer 1

X(dy/dx) + (2x - 1)y = (x - 1)

(dy/dx) + (2 - 1/x)y = (x - 1)/x

faktor integrasi :
e^∫(2 - 1/x dx)
= e^(2x - lnx)
= e^(lne^(2x) - lnx)
= e^(lne^(2x)/x)
= e^(2x)/x

e^(2x)/x((dy/dx) + (2 - 1/x)y = (x - 1)/x) = e^(2x)/x.(x - 1)/x
e^(2x)/x.y = ∫e^(2x)/x.(x - 1) dx
e^(2x)/x.y = ∫e^(2x) - e^(2x)/x dx
e^(2x)/x.y = (1/2)e^(2x) - ∫e^(2x)/x dx
y = (1/2)e^(2x)/(e^(2x)/x) - ∫e^(2x)/x dx / (e^(2x)/x)
y = x/2 - x.Ei(2x)/(e^(2x)) + C

==========

2xy(dy/dx) -y^2 + x^2 = 0
bagi dengan xy

2(dy/dx) - y/x + x/y = 0

misal, u = y/x
y = ux
dy/dx = u + x.du/dx

2(dy/dx) - y/x + x/y = 0
2(u + x.du/dx) - u + (1/u) = 0
2u + 2x.du/dx = u - 1/u
2x.du/dx = -u - 1/u
2du/(-u - 1/u) = dx/x
-2du/(u + 1/u) = dx/x
-2∫du/(u + 1/u) = ∫dx/x
-ln(u^2 + 1) = lnx + lnc
ln(u^2 + 1) = -(lnx + lnc)
ln(u^2 + 1) = -ln(xc)
ln(u^2 + 1) = ln(xc)^(-1)
u^2 + 1 = 1/(xc)
(y/x)^2 + 1 = 1/(xc)
(y/x)^2 = 1/(xc) - 1
y^2 = x/c - x^2

CMIIW