Finding the general solution of the differential equation XY '' + y '= 0

Finding the general solution of the differential equation XY '' + y '= 0

Solution 1: ∵ XY '' + y '= 0 = = > XDY' / DX = - y '
==>dy'/y'=-dx/x
==>Ln c3y 'C3 = - ln C3x + ln c3c1 (C1 is an integral constant)
==>y'=C1/x
==>Y = c1ln │ x │ + C2 (C2 is the integral constant)
The general solution of the original equation is y = c1ln │ x │ + C2 (C1, C2 are integral constants);
Solution 2: let t = ln │ x, then XY '= dy / DT, X & # 178; y' '= D & # 178; Y / dt & # 178; - dy / dt
Substituting into the original equation, we get D & # 178; Y / dt & # 178; - dy / dt + dy / dt = 0
==> d²y/dt²=0
==>Dy / dt = C1 (C1 is the integral constant)
==>Y = C1t + C2 (C2 is the integral constant)
==>y=C1ln│x│+C2
The general solution of the original equation is y = c1ln │ x │ + C2 (C1, C2 are integral constants)

What is the general solution of the differential equation y '= y + 1

dy/dx=y+1
dy/(y+1)=dx
The general solution is ln|y + 1| = x + C

The general solution of the differential equation (1 + x) y '+ 1 = 2E ^ (- y) is__________

(1+x)y'+1=2e^(-y)
(1+x)y'=2e^(-y)-1
dy/[2e^(-y)-1]=dx/(1+x)
e^ydy/[2-e^y]=dx/(1+x)
Integral is accessible - ln (2-e ^ y) + LNC = ln (1 + x)
Or: (1 + x) (2-e ^ y) = C