Separable Differential Equations

$y' = 12e^{3x}$.

Left: $12e^{3x}$. Right: $3\left(4e^{3x} - 2\right) + 6$.

Right: $12e^{3x} - 6 + 6 = 12e^{3x}$.

Both sides are $12e^{3x}$ for every $x$.

$y = (2 - x)^{-1}$, so $y' = (2 - x)^{-2} = \dfrac{1}{(2 - x)^2}$.

Left: $\dfrac{1}{(2 - x)^2}$. Right: $\left(\dfrac{1}{2 - x}\right)^2$.

Right: $\dfrac{1}{(2 - x)^2}$.

Equal for every $x \neq 2$.

$y' = 3x^2 + 1$.

Left: $3x^2 + 1$. Right: $\dfrac{3(x^3 + x)}{x}$.

Right: $3x^2 + 3$.

$3x^2 + 1 \neq 3x^2 + 3$.

$f(x) = 6x^2$, $g(y) = y^2$.

$\dfrac{dy}{y^2} = 6x^2\,dx$ (and $y = 0$ is a constant solution).

$-\dfrac{1}{y} = 2x^3 + C$.

$y = -\dfrac{1}{2x^3 + C}$.

$f(x) = \dfrac{1}{1 + x^2}$, $g(y) = y$.

$\dfrac{dy}{y} = \dfrac{dx}{1 + x^2}$.

$\log|y| = \arctan x + C$.

$|y| = e^Ce^{\arctan x}$, so $y = Ae^{\arctan x}$ ($A = 0$ gives the solution $y = 0$).

$f(x) = x$, $g(y) = 1 + y^2$ (never $0$).

$\dfrac{dy}{1 + y^2} = x\,dx$.

$\arctan y = \dfrac{x^2}{2} + C$.

$y = \tan\left(\dfrac{x^2}{2} + C\right)$.

$y\,dy = x\,dx$, so $\dfrac{y^2}{2} = \dfrac{x^2}{2} + C$.

$x = 0$, $y = -3$: $\dfrac{9}{2} = C$.

$y^2 = x^2 + 9$, and $y(0) < 0$, so $y = -\sqrt{x^2 + 9}$.

$y(0) = -\sqrt{9} = -3$.

$e^{x - y} = e^xe^{-y}$, so $e^y\,dy = e^x\,dx$ and $e^y = e^x + C$.

$e^1 = e^0 + C$, so $C = e - 1$.

$e^y = e^x + e - 1$, so $y = \log\left(e^x + e - 1\right)$.

$y(0) = \log(1 + e - 1) = \log e = 1$.

$\dfrac{dy}{y} = 3x^2\,dx$, so $\log|y| = x^3 + C$.

$\log 5 = 0 + C$, so $C = \log 5$.

$|y| = e^{\log 5}e^{x^3} = 5e^{x^3}$; $y(0) > 0$, so $y = 5e^{x^3}$.

$y(0) = 5e^0 = 5$.

$\dfrac{dP}{P(3 - P)} = dt$.

$\dfrac{1}{P(3 - P)} = \dfrac{1}{3}\left(\dfrac{1}{P} + \dfrac{1}{3 - P}\right)$.

$\dfrac{1}{3}\log\left|\dfrac{P}{3 - P}\right| = t + C$.

$P(0) = 1$ gives $\dfrac{P}{3 - P} = \dfrac{1}{2}e^{3t}$ (positive), so $P = \dfrac{3}{1 + 2e^{-3t}}$.

$e^{-3t} \to 0$, so $P \to 3$.

$\dfrac{dx}{(x - 1)(x - 4)} = dt$.

$\dfrac{1}{(x - 1)(x - 4)} = \dfrac{1}{3}\left(\dfrac{1}{x - 4} - \dfrac{1}{x - 1}\right)$.

$\dfrac{1}{3}\log\left|\dfrac{x - 4}{x - 1}\right| = t + C$.

At $x = 2$ the ratio is $-2$, so $\dfrac{x - 4}{x - 1} = -2e^{3t}$, giving $x = \dfrac{4 + 2e^{3t}}{1 + 2e^{3t}}$.

Divide by $e^{3t}$: $x = \dfrac{4e^{-3t} + 2}{e^{-3t} + 2} \to \dfrac{2}{2} = 1$.

$\dfrac{dQ}{dt} = -kQ$.

$\log|Q| = -kt + C$, so $Q = Ae^{-kt}$.

$Q(0) = 200$ gives $A = 200$. $Q(4) = 100$ gives $e^{-4k} = \tfrac{1}{2}$, so $k = \dfrac{\log 2}{4}$.

$25 = 200e^{-kt}$: $e^{-kt} = \tfrac{1}{8} = \left(\tfrac{1}{2}\right)^3$, so $t = 3 \times 4 = 12$.

Surface area is proportional to $V^{2/3}$, so $\dfrac{dV}{dt} = -kV^{2/3}$.

$V^{-2/3}\,dV = -k\,dt$, so $3V^{1/3} = -kt + C$.

$V(0) = 64$: $C = 12$. $V(6) = 27$: $9 = -6k + 12$, so $k = \tfrac{1}{2}$. Then $V^{1/3} = 4 - \dfrac{t}{6}$.

Gone when $4 - \dfrac{t}{6} = 0$: $t = 24$.

$\dfrac{dB}{dt} = 0.05B - 2000 = 0.05(B - 40{,}000)$.

$\dfrac{dB}{B - 40{,}000} = 0.05\,dt$, so $\log|B - 40{,}000| = 0.05t + C$.

$B(0) = 32{,}000$: $B - 40{,}000 = -8{,}000e^{0.05t}$. The rate $0.05$ is given, so there is no $k$ to find.

$B = 0$ when $e^{0.05t} = 5$: $t = 20\log 5 \approx 32.2$.