Question

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

Answer 1

Substitute
`v(x)=x-y(x)+1`, so that

`(\mathrm dv)/(\mathrm dx)=1-(\mathrm dy)/(\mathrm dx)`

Then the resulting ODE in
`v(x)` is separable, with

`1-(\mathrm dv)/(\mathrm dx)=v^2\implies(\mathrm dv)/(1-v^2)=\mathrm dx`

On the left, we can split into partial fractions:

`\frac12\left(\frac1{1-v}+\frac1{1+v}\right)\mathrm dv=\mathrm dx`

Integrating both sides gives

`\frac+\ln2=x+C`

`\frac12\ln|1-v^2|=x+C`

`1-v^2=e^(2x+C)`

`v=\pm\sqrt{1-Ce^(2x)}`

Now solve for
`y(x)`:

`x-y+1=\pm\sqrt{1-Ce^(2x)}`

`\boxed{y=x+1\pm\sqrt{1-Ce^(2x)}}`