Linear first-order differential equations: the integrating factor

Worked solutions

Problem 1

1. $\mu = e^{\int2\,dx} = e^{2x}$Already in standard form, with $P = 2$.
2. $(e^{2x}y)' = 6e^{2x}$Multiply both sides by $\mu$; the left side is now a product's derivative.
3. $e^{2x}y = 3e^{2x} + C$Integrate both sides.
4. $y = 3 + Ce^{-2x}$, and $y(0) = 3 + C = 1$ gives $C = -2$Divide by $\mu$, then use the initial condition.
5. $y = 3 - 2e^{-2x}$Check: $y' = 4e^{-2x}$ and $y' + 2y = 4e^{-2x} + 6 - 4e^{-2x} = 6$. The solution rises toward the equilibrium $y = 3$.

Problem 2

1. $P = -1$, so $\mu = e^{-x}$The sign of $P$ goes into the exponent.
2. $(e^{-x}y)' = e^{-x}e^{2x} = e^x$Multiply both sides by $e^{-x}$.
3. $e^{-x}y = e^x + C$, so $y = e^{2x} + Ce^x$Integrate, then divide by $e^{-x}$, which multiplies by $e^x$.
4. $y(0) = 1 + C = 0$ gives $C = -1$The initial condition.
5. $y = e^{2x} - e^x$Check: $y' - y = (2e^{2x} - e^x) - (e^{2x} - e^x) = e^{2x}$.

Problem 3

1. $y' + \dfrac2x\,y = x$Divide by $x$ to reach standard form; now $P = \tfrac2x$.
2. $\mu = e^{\int 2/x\,dx} = e^{2\ln x} = x^2$For $x > 0$, $\ln\lvert x\rvert = \ln x$.
3. $(x^2y)' = x^2\cdot x = x^3$Multiply the standard form by $x^2$. This is the original equation multiplied by $x$.
4. $x^2y = \dfrac{x^4}4 + C$, and $1 = \tfrac14 + C$ gives $C = \tfrac34$Integrate, then use $y(1) = 1$.
5. $y = \dfrac{x^2}4 + \dfrac{3}{4x^2}$Check: $xy' = \tfrac{x^2}2 - \tfrac{3}{2x^2}$ and $2y = \tfrac{x^2}2 + \tfrac{3}{2x^2}$, which add to $x^2$. For example $y(2) = 1 + \tfrac3{16} = \tfrac{19}{16}$.

Problem 4

1. $\mu = e^x$, so $(e^xy)' = xe^x$$P = 1$.
2. $\displaystyle\int xe^x\,dx = xe^x - e^x$By parts with $u = x$, $dv = e^x\,dx$.
3. $e^xy = (x - 1)e^x + C$, so $y = x - 1 + Ce^{-x}$Add the constant before dividing by $e^x$.
4. $y(0) = -1 + C = 0$ gives $C = 1$The initial condition.
5. $y = x - 1 + e^{-x}$Check: $y' + y = (1 - e^{-x}) + (x - 1 + e^{-x}) = x$. For large $x$ the solution follows the line $y = x - 1$, one unit behind the input.

Problem 5

1. $\displaystyle\int\tan x\,dx = -\ln\lvert\cos x\rvert$, so $\mu = e^{-\ln\cos x} = \dfrac{1}{\cos x} = \sec x$$\cos x > 0$ on this interval.
2. $(y\sec x)' = \cos x\sec x = 1$Multiply both sides by $\sec x$.
3. $y\sec x = x + C$, so $y = (x + C)\cos x$Integrate, then multiply by $\cos x$.
4. $y = (x + 1)\cos x$$y(0) = C = 1$. Check: $y' = \cos x - (x + 1)\sin x$ and $y\tan x = (x + 1)\sin x$, which add to $\cos x$.

Problem 6

1. $\mu = e^{\int2x\,dx} = e^{x^2}$, so $(e^{x^2}y)' = xe^{x^2}$$P = 2x$.
2. $e^{x^2}y = \tfrac12e^{x^2} + C$, so $y = \tfrac12 + Ce^{-x^2}$, and $C = -\tfrac12$$\int xe^{x^2}\,dx = \tfrac12e^{x^2}$ by $u = x^2$.
3. Separating: $y' = x(1 - 2y)$, so $\dfrac{dy}{1 - 2y} = x\,dx$ and $-\tfrac12\ln\lvert1 - 2y\rvert = \tfrac{x^2}2 + c$Here the right side factors, so this one is separable as well.
4. $1 - 2y = Ae^{-x^2}$ with $A = 1$ from $y(0) = 0$The same family.
5. $y = \tfrac12\big(1 - e^{-x^2}\big)$Both methods agree, as they must. With $P \ne 0$, an equation $y' + Py = Q$ is separable exactly when $\tfrac QP$ is a constant; here it is $\tfrac12$, the value $y$ settles at.

Problem 7

1. $Q' = 0.2\cdot5 - \dfrac{Q}{100}\cdot5 = 1 - \dfrac{Q}{20}$In: $1$ kg/min. Out: the concentration $\tfrac{Q}{100}$ times $5$ L/min. The volume stays at $100$ L.
2. $Q' + \dfrac{Q}{20} = 1$, $\mu = e^{t/20}$, $(e^{t/20}Q)' = e^{t/20}$Standard form with $P = \tfrac1{20}$.
3. $e^{t/20}Q = 20e^{t/20} + C$, so $Q = 20 + Ce^{-t/20}$, and $Q(0) = 10$ gives $C = -10$Integrate, divide, apply $Q(0)$.
4. $Q(20) = 20 - 10e^{-1}$; $Q = 19$ when $10e^{-t/20} = 1$, $t = 20\ln10$The gap to $20$ kg shrinks by a factor of $e$ every $20$ minutes.
5. $Q = 20 - 10e^{-t/20}$; $Q(20) \approx 16.32$ kg; $19$ kg at $t = 20\ln10 \approx 46.05$ minIn the long run the tank holds $100 \times 0.2 = 20$ kg, the inflow concentration.

Problem 8

1. $\mu = e^x$, $(e^xy)' = e^x\sin x$$P = 1$.
2. $\displaystyle\int e^x\sin x\,dx = \frac{e^x(\sin x - \cos x)}{2}$By parts twice, solving for the integral when it reappears.
3. $y = \dfrac{\sin x - \cos x}2 + Ce^{-x}$, and $y(0) = -\tfrac12 + C = 0$ gives $C = \tfrac12$Divide by $e^x$ and apply $y(0) = 0$.
4. $\dfrac{\sin x - \cos x}2 = \dfrac{1}{\sqrt2}\sin\Big(x - \dfrac\pi4\Big)$$\sin(x - \tfrac\pi4) = \tfrac{\sin x - \cos x}{\sqrt2}$.
5. $y = \tfrac12(\sin x - \cos x) + \tfrac12e^{-x}$; for large $x$, $y \approx \tfrac{1}{\sqrt2}\sin\big(x - \tfrac\pi4\big)$The transient $\tfrac12e^{-x}$ dies out. What is left follows the input with its amplitude cut to $\tfrac1{\sqrt2} \approx 0.707$ and a lag of $\tfrac\pi4$: smoothing always costs delay, which is the tradeoff in choosing an EMA's decay rate.

Problem 9

1. $V(t) = 20 + t$The volume grows by $2 - 1 = 1$ L/min.
2. $Q' = 3\cdot2 - \dfrac{Q}{20 + t}\cdot1 = 6 - \dfrac{Q}{20 + t}$The outflow concentration is $\tfrac{Q}{V(t)}$, with the current volume.
3. $\mu = e^{\int dt/(20 + t)} = 20 + t$, so $\big((20 + t)Q\big)' = 6(20 + t)$$P = \tfrac{1}{20 + t}$.
4. $(20 + t)Q = 3(20 + t)^2 + C$, and $Q(0) = 0$ gives $C = -1200$$\int6(20 + t)\,dt = 3(20 + t)^2$; at $t = 0$, $0 = 1200 + C$.
5. $Q = 3(20 + t) - \dfrac{1200}{20 + t}$; at $t = 20$, $90$ g in $40$ L, or $2.25$ g/L$Q(20) = 120 - 30$. The concentration approaches $3$ g/L, the inflow's, as the tank grows.

Problem 10

1. $v' = -\dfrac{y'}{y^2} = -\dfrac{y - y^2}{y^2} = 1 - v$A Bernoulli equation with $n = 2$, so $v = y^{1-2} = \tfrac1y$.
2. $v' + v = 1$, $\mu = e^x$, so $e^xv = e^x + C$ and $v = 1 + Ce^{-x}$Linear in $v$, with $P = 1$.
3. $v(0) = 2$ gives $C = 1$$v(0) = \tfrac{1}{y(0)}$.
4. $y = \dfrac{1}{1 + e^{-x}}$, the sigmoidCheck: $y' = \tfrac{e^{-x}}{(1 + e^{-x})^2} = y(1 - y)$. Separating with partial fractions gives the same answer; the substitution avoids the partial fractions.