Let $$v'= 0, 4608 + 0, 06242v − 0, 001717v^2$$ where,$v(0)=27.(7)$
I tried to solve this PVI and i got
$$v=\frac{6.293\cdot e^{\frac{x}{0.0243}+c}+42.647}{1-e^{\frac{x}{0.0243}+c}}$$
Applying $v(0)=27.(7)$ I've reached,
C=$\ln(-0.436420)$, which is quite impossible.
Can someone explain to me what this means?Are my calculations correct? Thanks for helping, have a nice day!
On
You missed to apply an absolute value to a logarithm at some point, $\int\frac{du}u=\ln|u|+c$. If you did this correctly, and resolved this by an unknown sign of the exponential, then using $C=\pm e^c$ gives the general solution $$ v(x)=\frac{6.293\cdot Ce^{\frac{x}{0.0243}}+42.647}{1-Ce^{\frac{x}{0.0243}}} $$ Now inserting $v(0)=27.77\bar7$ is just an arithmetic problem, giving, if your numbers are right, $C=−0.436420$. This has the consequence that there will be no poles anywhere. This compatible with the fact that $v(0)$ is inside the interval $[-6.29293532,\,42.64704132]$ between the roots of the right side, forcing the solution to be bounded and thus to exist for all $x$.
Assuming your solution to the ODE is correct, this implies there are no real solutions to the IVP.
Not sure how you arrived at your solution. One way to approach this is is to use separation of variables: $$ \begin{split} \frac{dv}{dt} &= av^2 + bv + c \\ dt &= \frac{dv}{av^2 + bv + c} \\ t + C &= \frac{1}{a} \int \frac{dv}{v^2 + (b/a) v + (c/a)} \end{split} $$ and you can complete the square. The other way is to factor the polynomial, if possible, to get $$ t + C = \frac{1}{a} \int \frac{dv}{(v-s)(v-t)} $$ for some $s,t \in \mathbb{R}$ and use partial fractions...