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

Question

asked Oct 4, 2020 162k views

1 Answer

Belky · Answer 1 · 2020-10-08T22:00:37+0000

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) :

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

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