Differential equation substitution f(x/t)

Question

Hello I have the differential equation $$x'=\frac{8t+10x}{17t+x}$$ Brought it in a eqation where I can substitute $$ u=\frac{x}{t}$$ and after some transformations I got the equation $$ \frac{17+u}{8-7u-u^2}du=\frac{1}{t}dt $$ with the following equation $$ -2ln(u-1)+ln(u+8)=ln(t)+c$$ Then I tried to multiply it with $$e$$ to make the $$ln()$$ disappear then I got the equation $$ \frac{u+8}{(u-1)^2}=te^c$$ but now I dont know how to resubstitute or to break the fracture.Thanks,Ciwan.

Answer 1

Yes, you have solved it. But usually,one does not look for singular solution(s) of such an ODE. Put $u=x/t$ in the ODE but treat $u$ as constant to get $u^2+7u-8=0$ and $u=-8,1$. So $x=-8t$ and $t$ are the singular solutions of thisODEe which are without any constant of integration. These two are additional solutions apart from the one obtained by the proposer which is the general solution with a constant $c$. You may see:

Can a first order homogeneous ODE have singular solution(s) (free of a constant)

Answer 2

$$-2ln(u-1)+ln(u+8)=ln(t)+c \implies \frac {u+8}{(u-1)^2} =ct $$

Go from there with $$u=x/t$$.