Question

solve the differential equation (y+y2)dx+xydy=0​

Answer 1

The solved equation is log(xy+x) + C = 0 for the differential equation (y + y²)dx + xydy = 0.

Given that,

We have to solve the differential equation (y + y²)dx + xydy = 0

We know that,

Take the equation

(y + y²)dx + xydy = 0

(y + y²)dx = -xydy

$\frac{1}{x}$ dx = $\frac{-y}{y+y^2}$ dy

Taking integration on both sides.

$\int \frac{1}{x} \, dx$ = $\int{\frac{-y}{y+y^2}} \, dy$

We know that $\int \frac{1}{x} \, dx$ = logx + C

logx +C = $\int{\frac{-1}{1+y}} \, dy$

logx + C = -log(y+1)

logx + log(y+1) +C =0

From logarithm formula loga + logb = log(ab)

log(x(y+1)) + C =0

log(xy+x) + C = 0

Therefore, The solved equation is log(xy+x) + C = 0

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