Integrate $\frac{dR}{dt}=\frac{t^3+R^3}{t^2R+tR^2}$ and find the solution for $R(1) = -1$

Question

Yesterday I asked for some help regarding this question which I found from here but I'm stuck again and I'll show you my working so far with my homework

First I got this equation separated successfully through help found here and got

$$\int \frac{u+u^2}{1-u^2}du=\int \frac{1}{t}dt$$ assuming $R=ut$

I then integrated the equation and got $$-u+1-\ln(u-1)=\ln(t)+c$$ by substitution of $v=u-1$ and $du=dv$

Replaced $u$ with $R$ again and got $$\frac{-R}{t}+1-\ln\left(\frac{R}{t}-1\right)=\ln(t)+c$$

finally I just have to use $R(1) = -1$ to find $c$ but I'm getting a math error and can't see where did I go wrong. Any form of hint will be much appreciated thank you

Answer 1

I assume your math error is $\ln(R/t-1)=\ln(-2)$. You really should have $\ln|u-1|$, not $\ln(u-1)$. Note that the "integration constant" is also allowed to have different values either side of $u=1$; it's what we call a locally constant function.

Answer 2

$$\int \frac{u+u^2}{1-u^2}du=\int \frac{1}{t}dt$$ $$\int \frac{u}{1-u}du=\int \frac{1}{t}dt$$ $$\int \frac{u-1+1}{1-u}du=\int \frac{1}{t}dt$$ $$-\int du-\int \frac{1}{u-1}du=\int \frac{1}{t}dt$$ $$-u-\ln |{u-1}|=\ln |{t}|+C_1$$ $$\dfrac R t+\ln |{R-t}|=C$$ Apply initial condition: $$R(1)=-1 \implies C=\ln 2-1$$ $$\dfrac R t+\ln |{R-t}|=\ln 2-1$$

Answer 3

$R(t)=tu(t)$ and $R(1)=-1$ implies that $u(1)=-1$ and

$$t\dot{u}+u = \frac{1+u^3}{u+u^2}=\frac{1-u+u^2}{u}$$

that is

$$ \frac{u}{1-u} \dot{u} = \frac{\dot{u}}{1-u} - \dot{u}=\frac{1}{t} $$

Integration over $[1,t]$

$$-\ln(1-u)+\ln(1-u(1)) - u+u(1)=\ln(t)$$

this means that $$-\log(1-u)-u=\ln(t)+1-\ln(2)$$

In terms of the original function $R$,

$\frac{R}{t}+\ln\Big(1-\frac{R}{t}\Big) =-\ln(t)-1+\ln(2)$