How to solve $x'=\frac{1-x-t}{x+t}$?

Question

I would be greatful if you help me to find the solutions of this differential equation:

$[E]: x'=\frac{1-x-t}{x+t}$.

The exercise gives a hint: use $u=ax+bt+c$. I am trying with $u=x+t$ and $u=1-x-t$ but I can't isolate $u$.

Thanks!

Answer 1

$$x' =\frac{1-x-t}{x+t} = \frac{1}{x+t} - 1 \implies x'+1 = \frac{1}{x+t}$$

Let $x+t = u \implies x'+1 = u'$

So, $$u' = \frac{1}{u}$$

Answer 2

Let $u=x+t$, then $u'=x'+1$ or $x'=u'-1$, thus the DE becomes $$u'-1=\frac{1-u}{u}$$ or $$u'=\frac{1}{u}$$ i.e. $$\frac{1}{2}u^2=t+c\\ \implies (x+t)^2=2t+2c $$

Answer 3

$$x'=\frac{1-x-t}{x+t}$$

$$x'+1=\frac{1-x-t}{x+t}+1$$ $$(x+t)'=\frac{1-x-t}{x+t}+1$$ $$(x+t)'=\frac{1}{x+t}$$ This is a separable DE. $$(x+t)d(x+t)=dt$$

Answer 4

$$ \frac{dx}{dt} = \frac{1-(x+t)}{x+t} $$ Let $$k = x+t$$ Then $$ \frac{dx}{dt} + 1 = \frac{dk}{dt}$$ $$\frac{dx}{dt} = \frac{dk}{dt} -1 $$and$$ \frac{dk}{dt} - 1 = \frac{1-k}{k} $$ $$ \frac{dk}{dt} = \frac{1-k}{k} + 1 = \frac{1}{k}$$ $$\int k\ dk = \int dt $$ $$\frac{k^2}{2} + c = t$$ substituting $k$ with $x+t$ $$\frac{(x+t)^2}{2} + c = t$$