solve differential equation (4-x)dy +2ydx = 0

Question

Find the general solution of the differential equation

Answer 1

 

Solve (1+e^x/y)dx+e^x/y(1-x/y)dy=0

Answer 2

(6+x)dy-2ydx=0

Answer 3

divide both sides by ((2*y) and (4-x))

(1/2*y)dy + (1/(4-x))dx = 0

take integral for both sides, 

(1/2 * ln y) - ln|4-x| = c

Answer 4

dy/dx=x^2/y(1+x^3)

Answer 5

(4-x)dy+2ydx=0 this is a separable first order differential equation

(4-x)dy=-2ydx

dy/y=2dx/(x-4) By integrating both sides;

ʃdy/y=2ʃdx/(x-4), we get

lny=2ln(x-4)+lnC

Then, y=ß(x-4)² ; With ß=lnC ( constant)

Answer 6

2ydx=(1+x)dy

Answer 7

((4-x)dy + 2ydx =0)1/(4-x)

dy + 2y/(4-y) =0

e^§2/(4-x) dx

e^(ln(4-x)^2)

e^(ln(4-x)^2) (dy + 2y/(4-y) =0)

e^(ln(4-x)^2) (y) = 0

(4-x)^2 (y) = 0

16y -8xy +x^2y = C

Is it correct??

 

 

 

Answer 8

lny=2ln(4-x)

Answer 9

dY/dx=x^3+y^3/x^2y+xy^2

Answer 10

anik

Answer 11

ljgcv

Answer 12

SOLVE    Dy\dx + x sine(2y) = x3 cos2y

Answer 13

show that M(X,Y) + N(x,y)=0 has an integration factor f(x) and an integration factor g(x) . then the equation is separable.

Answer 14

Reduce the second order differential equation into first order DE

D, where x, k, q, D are constant  

Answer 15

Reduce the second order differential equation into first order DE

D d2/dz2 = qsx/s+k  where x, k, q, D are constant

Answer 16

(4-x)dy+2ydx=0, (4-x≠0, y≠0)   Divide both sides by 2y(4-x), getting

(1/2y)dy+(1/(4-x))dx=0  

This is a separable first order differential equation, so the equation can be integrated as:

∫(1/2y)dy+∫(1/(4-x))dx=C  

Here,∫(1/2y)dx=½ln(y), ∫(1/(4-x)dx=-ln(4-x).   So, the equation can be rewritten as: 

½ln(y)-ln(4-x)=C   Use quotient rule of logarithm,getting

ln(y^½ /(4-x))=C   So,we have: y^½ /(4-x)=e^C=m, (m:constant).   Square both sides:

y/(4-x)²=m²=k, (k:constant).   Therefore, y=k(4-x)²

The answer is: y=k(4-x)²

CK: y=k(4-x)² ⇒ dy/dx=-2k(4-x), k=y/(4-x)², so dy/dx={-2y/(4-x)²}(4-x) ⇒

dy/dx=-2y/(4-x) ⇒ (4-x)dy=-2ydx ⇒ (4-x)dy+2ydx=0   CKD.