(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.