1.
xy2(xy+1)dx+x(xy−1)(xy+1)dy=0
x=0y=0xy+1=0=>y=−x1 Or
y2dx+(xy−1)dy=0P(x,y)=y2,Q(x,y)=xy−1
ðyðP=2y,ðxðQ=y,ðyðP=ðxðQðyðP−ðxðQ=2y−y=yμ=μ(y)
μ1dydμ=−P1(ðyðP−ðxðQ)
μ1dydμ=−y21(2y−y)
μdμ=−ydy
∫μdμ=∫(−ydy)=>ln∣μ∣=ln∣y1∣+lnC We choose
μ=y1 Multiplying the original differential equation by μ=1/y produces the exact equation:
ydx+(x−y1)dy=0 Indeed, now we have
ðyðP=1,ðxðQ=1,ðyðP=ðxðQ Solve the resulting equation. The function u(x,y) can be found from the system of equations:
{ðxðu=yðyðu=x−y1 It follows from the first equation that
u=∫ydx=xy+φ(y) Substitute this in the second equation to determine φ:
ðyðu=x+dydφ=x−y1
∫dφ=∫(−y1)dy=>φ(y)=ln(∣y∣1) Thus, the original differential equation has the following solutions:
x=0y=0y=−x1xy+ln(∣y∣1)=C1 where C1 is a constant.
2.
(x2+y2−y)dx+xdy=0 We can make sure that this equation is not exact:
P(x,y)=x2+y2−y,Q(x,y)=x
ðyðP=2y−1,ðxðQ=1,ðyðP=ðxðQ The difference of the partial derivatives is
ðyðP−ðxðQ=2y−1−1=2y−2 Using the integrating factor μ=x2+y2 we find that
ðyðz=2y,ðxðz=2x Calculate the following expression:
Qðxðz−Pðyðz=x(2x)−(x2+y2−y)(2y)==2x2+2y2−2y(x2+y2)=(x2+y2)(2−2y) As a result, we obtain the differential equation for the function μ(z):
μ1dzdμ=(Qðxðz−Pðyðz)1(ðyðP−ðxðQ)=
=(x2+y2)(2−2y)1(2y−2)=−x2+y21=−z1
By integrating we get the function μ(z):
∫μdμ=∫(−zdz)=>ln∣μ∣=−ln∣z∣+lnC We choose
μ=z1We can choose the integrating factor
μ=−z1=−x2+y21
−x2+y21(x2+y2−y)dx−x2+y21xdy=0 Or
(1−x2+y2y)dx+x2+y2xdy=0 The general solution u(x,y)=C is defined by the following system of equations:
{ðxðu=1−x2+y2yðyðu=x2+y2x Integrating the first equation with respect to x produces:
u(x,y)=∫(1−x2+y2y)dx=x−y∫x2+y21dx=
=x−arctan(yx)+φ(y) Substituting this in the second equation, we have:
ðyðu=x2+y2x+dydφ=x2+y2x
dydφ=0 Thus, the general solution of differential equation in implicit form is defined by the formula:
x−arctan(yx)=C where C is a constant.
3.
dxdy+y=3y3exy=0 Or
y−3dxdy+y−2=3ex The associated homogeneous differential equation is
y−3dxdy+y−2=0
dxdy+y=0
ydy=−dxy=C1e−x
dxdy=dxdC1e−x−C1e−x Substitute
dxdC1e−x−C1e−x+C1e−x=3C13e−3xex
dxdC1=3C13e−x
C13dC1=3e−xdx
∫C13dC1=∫3e−xdx
−2C121=−3e−xdx−2C2
C1=6e−xdx+C21 Thus, the original differential equation has the following solutions:
y=0y=ex6e−xdx+C21 where C1 is a constant.
Comments