ECUACIONES DIFERENCIALES

PRESENTADO POR
CRISTIAN RAMIREZ OLARTE
CARLOS BOVEA ISAZA

CORPORACION UNIVERSITARIA DE LA COSTA
CUC
IV SEMESTRE DE INGENIERIA INDUSTRIAL
GRUPO AN
BARRANQUILLA - ATLANTICO

dy/dx=5y

dy/dx-5y=0 P_((x))=-5 e^∫▒〖-5dx〗=e^(-5x) → d/dx [e^(-5x).y]=e^(-5x) (0) d/dx e^(-5x) y=0

e^(-5x) y=C y=Ce^5x

dy/dx+2y=0 P_((x))=2 e^(∫2dx)=e^2x → d/dx [e^2x y]=e^2x (0) e^2x y=C y=Ce^(-2x)

dy/dx+y=e^3x

P_((x) )=1 e^(∫dx)=e^x

d/dx [e^x y]=e^3x e^x→de^x y=e^4x dx

e^x y=1/4 e^4x+C

y=1/4 e^3x+Ce^(-x) 3 dy/dx+12y=4

dy/dx+4y=4/3

P_((x) )=4 e^(∫4dx)=e^4x

d/dx [e^4x y]=4/3 e^4x …ver más…

y^'+(1+x)y=e^(-x) sen 2x dy/dx+((1+x)/x)y=(e^(-x) sen2x)/x P_((x))=(1+x)/x e^(∫(1+x)/x dx)=e^(∫dx/x+∫dx)=e^(lnx+x)=e^lnx e^x=xe^x d/dx [xe^x y]=(xe^x e^(-x) sen2x)/x xe^x y=-1/2 cos2x+C y=Cx^(-1) e^(-x)-1/2 x^(-1) e^(-x) cos2x

ydx-4(x+y^6 )dy=0 dx/dy-(4 (x+y^6))/y=0 dx/dy-4x/y-4y^5=0 dx/dy-(4 (x+y^6 ))/y-→P_((x) )=-4/y e^(-∫4/y dy = e^(-4lny) )=e^(lny-4)=y^(-4) d/dy [y^(-4) x]=4y^3.y^(-4) -→ xy^(-4)=4ydy x/y^4 =(4y^2)/2+C x=2y^6+Cy^4 ydx=(ye^y-2x)dy ydx=ye^y dy-2xdy ÷ y dx=e^y dy-2x/y dy÷ dy dx/dy=e^y-2x/y -→ dx/dy+2/y x=e^y P_((y) )=2/y -→ e^(∫2 dy/y)=e^2lny=e^(lny^2 )=y^2 d/dy [y^2 x]=y^2 e^y -→y^2 x= ∫y^2 e^y d y^2 x=y^2 e^y-∫e^y 2ydy =y^2 e^y-2∫e^y ydy =y^2 e^y-2[e^y y-∫e^(y ) dy] y^2 x=y^2 e^y-2e^y y+2e^y+C x=e^y-(2e^y)/y+(2e^y)/y^2 +Cy^(-2)

cosx dy/dx+(senx)y=1 ÷cos⁡x dy/dx+(tang x)y=sec⁡〖x 〗 P_((x))=tang x e^(∫tang x dx)=e^ln⁡〖(sec⁡〖x)〗 〗 =sec⁡x d/dx [y sec x]=〖sec〗^2 x y sec⁡〖x=∫〖sec〗^2 x dx〗 y sec⁡〖x=tang x+C〗 y=(tang x)/sec⁡x +C/sec⁡x =(((sen x)/cos⁡x )/(1/cos⁡x ))+C/(1/cos⁡x