Question

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

Answer 1

It looks like the differential equation is

`y'-\frac y{x+1}=2`

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

`(y')/(x+1)-\frac y{(x+1)^2} = \frac2{x+1}`

The left side is now a derivative of a product,

`\left(\frac y{x+1}\right)' = \frac2{x+1}`

Integrate both sides with respect to x :

`\frac y{x+1} = 2\ln|x+1|+C`

Solve for y :

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