How to differentiate independent variable

Question

I have the equation $v' = f(t, v) = 25(\cos t - v + t/5)$, with initial conditions $v(0) = 1$. How can I evaluate $\frac{df(t,v)}{dv}$ at $t = 0, v = 1$? In particular, what is $dt/dv$ when computing $\frac{df(t,v)}{dv}$? Normally I would say it is 0, but $v$ is a function of $t$ so I'm not so sure. I also tried using $dt/dv = 1/(dv/dt) = 1/v'$, but that is undefined at $t = 0, v = 1$.

Answer 1

If you want to find solution of the equation, you can invoke superposition principle and solve homogeneous equation via, say, separation of variables. Adding a particular solution for non-homogeneous problem to the general solution of homogeneous will give us final answer.

$$ \left\lbrace\begin{aligned} v' &= -25 v + 5 t + 25\cos t\\ v(0)&=1 \end{aligned}\right. \qquad\xrightarrow[\text{problem}]{\text{homogenious }}\qquad \left\lbrace\begin{aligned} v' &= -25 v\\ v(0)&=1 \end{aligned}\right. \implies v_{h}\left(t\right) = e^{-25t} $$ Since non homogeneous RHS involves $t$ and trig functions we guess ansatz for a particular solution as $\,\overline{v}(t) = a\sin t + b\cos t + ct$. Substituting ansatz into original equation we get

$$ \left\lbrace\begin{aligned} \overline{v}(t) &= a\sin t + b\cos t + ct +d \\ \overline{v}'(t) &= a\cos t - b\sin t + c \end{aligned}\right. \implies \\ \implies a\cos t - b\sin t + c = 5t + 25\cos t -25\left(a\sin t +b\cos t +ct+d\right) $$ Collecting coefficients in front of alike terms we get $$ \left.\begin{alignedat}{4} \left(a -25 + 25b \right)\cos t &=0 &\implies a&=25\left(1-b\right)\\ \left(25 a -b \right)\sin t &= 0&\implies b &= 25a \end{alignedat}\right\rbrace \implies a = 25-625a \implies \left\lbrace\begin{aligned} a &= 25/626 \\ b &= 625/626 \end{aligned}\right. \\\left(5 -25c\right)t = 0 \implies c = \frac{1}{5}, \qquad d =0 $$ Therefore the final solution of the original problem $v\left(t\right) = v_h\left(t\right)$ takes form $$\boxed{ v\left(t\right) = e^{-25t} + \frac{1}{5}t + \frac{25}{626}\big(\sin t + 25\cos t\big)} $$

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The formulation of your question suggests that you are trying to investigate phase plane of the equation for stable/unstable nodes. In this case you treat $f\left(t,v\right)$ as a function of two independent variables:

$$ f\left(t,v\right) = -25 v + 5 t + 25\cos t \left.\right.\implies \left\lbrace\begin{aligned} f_t\left(t,v\right)\big\rvert_{\substack{t=0\\v=1}} &= 5 - 25\sin t\big\rvert_{\substack{t=0\\v=1}} = 5 \\ \left.\right.f_v\left(t,v\right)\big\rvert_{\substack{t=0\\v=1}} &= -25 \end{aligned}\right. $$