Ejercicios De Ecuaciones Diferenciales

ECUACIONES DIFERENCIALES ORDINARIAS HOMOGENEAS

Resolver
4x2+xy-3y2dx+(-5x2+2xy+y2)dy=0
Cv y=ux → dy=udx+xdu
4x2+x2u-3ux2dx+-5x2+2x2u+ux2udx+xdu=0
4x2+x2u-3ux2dx+-5ux2+2ux2+u3x2dx+(-5x3+2x3u+u2x3)du=0
4x2-4x2u-ux2+u3x2dx+(-5x3+2x3u+(ux)2)du=0
x24-4u-u2+u3dx+ x3-5+2u+u2du=0
1xdx+u2+2u-5u3-u2-4u+4du=0
lnx+23lnyx-1+34lnyx-2-512lnyx+2=cxcosecyx-ydx+xdy=0
Cv y=ux → dy=udx+xdu
xcosecu-uxdx+x(udx+xdu)=0
xcosecudx-uxdx+uxdx+x2du=0
1xdx+1cosec(u)du=0
lnx-cos⁡(yx)=c

x2dydx=4x2+7xy+2y2
Cv y=ux → dy=udx+xdu
x2dy=4x2+7xy+2y2dx
x2udx+xdu=(4x2+7xux+2(u2x2))=dx
ux2dx+x3du=4x2+7ux2+2u2x2dx
ux2dx+x3du=4x2dx+7ux2dx+2u2x2dx
x3du=4x2dx+6ux2dx+2u2x2dx
x3du=x2+4+6u+2u2dx
xdu=2u2+6u+4dxxdu2=u2+3u+2dx
duu2+3u+2=2dxx
x2y+2x=c(y+x)

yx2+xy-2y2dx+x3y2-xy-x2dy=0
Cv y=ux → dy=udx+xdu
yxx2+xy-2y2dx+(3y2-xy-x2dy=0
ux2+ux2-2u2x2dx+3u2x2-ux2-x2udx+xdu=0
ux2+u2x2-2u3x2)dx+3(u3x2-ux2-ux2)dx+(3u2x3-ux3-x3)du=0
u3x2dx+3u2x3-ux3-x3du=0
(u3x2)dx+x33u2-u-1du=0
1xdx+31u-1u2-1u3du=0
lnx+3lnu+1u+121u2=c
lnx+lnu3+xy+12xy2=c //y
y2lny3x2++xy+12x2=cy2 //22y2lny3x2+2xy+2x2=cy2

xdydx=y2-x2
Cv y=ux → dy=udx+xdu
xdy=y2-x2dx
xudx+xdu=(ux2-x2dx
udx+x2du=xu2-1dx
x2du=xu2-1-udx
duu2-1-u= dxx
y+y2-x2=cx3ey(y+y2-x2x2

x+(x-y)eyx)dx+xeyxdy=0
Cv y=ux → dy=udx+xdu
xx-uxeudx+xeuudx+xdu=0
x+xeu-uxeudx+uxeudx+x2eudu=0
x1+eudx+x2eudu=0
1xdx+eu1+eudu=0
x1+ex=k

dydx=yx+sinyx
Cv y=ux →dy=udx+xdu
dy=(yx+sinyxdx
udx+xdu=(u+sinu)dx
udx+xdu=udx+sinudx
xdu-sinudx=0
dusinu-dxx=0
cscudu-dxx=0
lncscu-cotu-lnx=lnc
lncscu-cotu=lnx+lnc
lncscu-cotu=lnxc
cscyx-cotyx=xc

xsin(yx)dydx=ysinyx+x
Cv y=ux → dy=udx+xdu
xsin(yx)dy-(ysinxy+x)dx=0
xsinuudx+xdu-(uxsinu+x)dx=0
uxsinudx+x2sinu du-(uxsinu+x)dx=0
x2sinu du-xdx=0
sinudu-1xdx=0
-cosu-lnx-lnc=0cosu+lnx+lnc=0
cosyx+lncx=0
BERMOULLI
x2ydx+3x4-y3dy=0
x2ydxdy+3x4-y3=0 //1x2y
dxdy+31yx2=y2x-2 //x2
x2dxdy+3x4y=y2 //3
3x2dxdy+9x4y= 3y2
cv z=x3 → dzdx=3x2dxdy
dzdx+31yz43=3y2
15x4y12=4y15+c

xdydx-y=xkyn //1x
dydx-x-1y=xk-1yn //y-n
y-ndydx-x-1y1-n=xk-1 //(1-n)
1-ny-ndydx-1-nx-1y1+n=(1-n)xk-1
cv z=y1-ndzdx=(1-n)y-ndydx
dzdx-1-nx-1z=(1-n)xk-1
z=e--1-nx-1dx(e-1-nx-1dx1-nxk-1dx+c)
z=e1-nlnx(e-1-nlnx1-nxk-1dx+c
z=x1-n(xn-11-nxk-1dx+c
z= x1-n1-nxn+k+3dx+c
z=x1-n1-nxn+k-1n+k-1+c
n+K-1y1-n=1-nxk+cx1-n+cx1-n n≠1,K+n≠1
xk=Klncyx,n=1,K≠0;y=cx2,n=1,K=0
y1-n=(1-n)x1-nlncx, n≠1,K+n=1

x2+x+1ydydx+2x+1y2=2x-1
x2+x+1dydx+2x+1y=(2x-1)y-1 //n=-1
dydx+2x+1x2+x+1y=(2x-1x2+x+1)y-1//y
ydydx+2x+1x2+x+1y2=2x-1x2+x+1 //2
2ydydx+22x+1x2+x+1y2=22x-1x2+x+1
C.V
z=y2 dzdx=2ydydx
dzdx+22x+1x2+x+1z=22x-1x2+x+1
z=e-22x+1x2+x+1dx(e22x+1x2+x+1dx22x-1x2+x+1dx+c)
z=e-2lnx2+x+1(e2lnx2+x+122x-1x2+x+1 dx+c)
z=elnx2+x+12((x2+x+1)2 22x-1x2+x+1 dx+c)
z= (x2+x+1)-22(2x3+x2+x-1dx+c)
z= (x2+x+1)-2(212x4+x33+x22-x+c)
z= (x2+x+1)-2(x4+23x3+x2-2x+c)3y2=3x4+2x3+3x2-6x+kx2+x+12
3y2x2+x+12= 3x4+2x3+3x2-6x+k

xy2+x2y2+3dx+x2ydy=0
x2ydydx+xy2+x2y2+3=0
x2ydydx+xy21+x+3=0 //x-2y-1
dydx+1+xxy=-3x-2y-1 //2y
2ydydx+1+xxy2=-6x2
Cv z=y2 → dzdx=2ydydx
dzdx+1+xxz=-6x2
z=e-(1+xx)dxe1+xxdx-6x2dx+c
z=e-lnx-xelnx+x-6x2dx+c
z=x-1ex -6x3exdx+c
z=x-1ex -6(x3ex-3x2exdx)+c
z=x-1ex -6x3ex+18(x2ex-2xexdx)+cz=x-1ex -6x3ex+18x2ex-36exx-1+c
z=x-1 -6x3+18x2-36x-1+ce-x
y2=x-1 -6x3+18x2-36x-1+ce-x

ECUACIONES DIFERENCIALES DE RICATTI
dydx=y2-1xy+1-14x2, una solucion es φx=12x+tgx
dydx=-1xy+y2+1-14x2
Si y=12x+tgx+z → dydx=-12x2+sec2x+dzdx
-12x2+sec2x+dzdx=12x+tgx+z2-1x12x+tgx+z+1-14x2
sec2x+dzdx=14x2+tg2x+z2+tgxx+zx+2ztgx-12x2-tgxx-zx+1+14x2
sec2x+dzdx=z2+2ztgx+1...