Question

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

Answer 1

The required answer is
`x+C=(1)/(2)\ln|(2+x-y)/(y-x)|`.

Explanation:

The given differential equation is

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

Substitute u=x-y+1 in the above equation.

`(du)/(dx)=1-(dy)/(dx)`

`(dy)/(dx)=1-(du)/(dx)`

`1-(du)/(dx)=u^2`

`1-u^2=(du)/(dx)`

Using variable separable method, we get

`dx=(du)/(1-u^2)`

Integrate both the sides.

`\int dx=\int (du)/(1-u^2)`

`x+C=(1)/(2)\ln|(1+u)/(1-u)|`
`[\because \int (dx)/(a^2-x^2)=(1)/(2a)\\|(a+x)/(a-x)|+C]`

Substitute u=x-y+1 in the above equation.

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

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

Therefore the required answer is
`x+C=(1)/(2)\ln|(2+x-y)/(y-x)|`.