Question

Giải các phương trình vi phân sau y' - y/x+1 = 2

Answer 1

It looks like the differential equation is

$y'-\dfrac y{x+1}=2$

Multiply both sides by 1/(x + 1) :

$\dfrac{y'}{x+1}-\dfrac y{(x+1)^2} = \dfrac2{x+1}$

The left side is now a derivative of a product,

$\left(\dfrac y{x+1}\right)' = \dfrac2{x+1}$

Integrate both sides with respect to x :

$\dfrac y{x+1} = 2\ln|x+1|+C$

Solve for y :

$y = 2(x+1)\ln|x+1| + C(x+1)$