first order linear PDE solving

Question

$$\dfrac{\partial{\phi}}{\partial{i}}=0$$

$$\dfrac{\partial{\phi}}{\partial{v}}=E-v-i R_0$$ Where E,$R_0$ are constants. How do I solve these kind of PDE's.

Answer 1

I suppose that $i$ is a variable and not the square root of $-1$. On the one hand ${∂_v}{∂_i}ϕ=0$ and on the other ${∂_i}{∂_v}ϕ=-R_0$. So, if the two are equal, $R_0$ must vanish in order to avoid a contradiction.

Answer 2

$$\dfrac{\partial{\phi}}{\partial{i}}=0$$

$$\dfrac{\partial{\phi}}{\partial{v}}=E-v-i R_0$$ Where E,$R_0$ are constants.

Then

(1) $\dfrac{\partial{\phi}}{\partial{i}}=0$ => ${\phi}=f(v)$

(2) $\dfrac{\partial{\phi}}{\partial{v}}=E-v-i R_0$ => ${\phi}=Ev-\dfrac{1}{2}v^2-i R_0v + f(i)$

Taking partial of (2) wrt $i$ yields:

(3) $\dfrac{\partial{\phi}}{\partial{i}}= - R_0v + f'(i) = 0$ => $f'(i) = R_0v$ => $f(i) = R_0vi + K$

Plugging (3) into (2) yields:

${\phi}=Ev-\dfrac{1}{2}v^2-i R_0v + f(i) $ => ${\phi}=Ev-\dfrac{1}{2}v^2+ K$, where $K$ is a constant