Exercises — First-order linear ODEs — integrating factor method (derivation)

(a) Linear. $y$ and $\frac{dy}{dx}$ appear only to the first power, never multiplied together. Already standard (coefficient of $\frac{dy}{dx}$ is $1$): $P=3,Q=x^2$. (b) NOT linear. The term $y^2$ raises $y$ to the second power — the definition forbids any power of $y$ other than the first. (c) Linear. But coefficient of $\frac{dy}{dx}$ is $x$, not $1$. Divide by $x$: $\frac{dy}{dx}+\frac{1}{x}y=\frac{\sin x}{x}$, so $P=\frac{1}{x},Q=\frac{\sin x}{x}$. (d) NOT linear. $\frac{dy}{dx}$ sits inside $\sin (\cdot )$ — a function of the derivative. Linear means $y$ and $y^{\prime }$ appear bare, to the first power, never inside another function.

(a) $\mu =e^{\int 5dx}=e^{5x}$. (b) $\mu =e^{\int \frac{2}{x}dx}=e^{2\ln |x|}=|x{|}^2=x^2$ (take $x^2$ on our working domain). (c) Here $P=-\tan x$, so $\mu =e^{\int -\tan xdx}$. Since $\int \tan xdx=-\ln |\cos x|$, we get $\int -\tan xdx=\ln |\cos x|$, hence $\mu =e^{\ln |\cos x|}=|\cos x|=\cos x$ (on an interval where $\cos x>0$).

Standardize: already standard, $P=3,Q=6$. Mu: $\mu =e^{\int 3dx}=e^{3x}$. Multiply & collapse: $\frac{d}{dx}(e^{3x}y)=6e^{3x}$. Integrate: $e^{3x}y=\int 6e^{3x}dx=2e^{3x}+C$. Divide: $y=2+C{e}^{-3x}$.

Standardize: already standard, $P=\frac{1}{x},Q=3x$. Mu: $\mu =e^{\int \frac{1}{x}dx}=e^{\ln x}=x$. Multiply & collapse: $\frac{d}{dx}(xy)=x\cdot 3x=3x^2$. Integrate: $xy=\int 3x^{2}dx=x^3+C$. Divide: $y=x^2+\frac{C}{x}$.

Standardize: already standard, $P=2x,Q=x$. Mu: $\mu =e^{\int 2xdx}=e^{x^2}$. Multiply & collapse: $\frac{d}{dx}(e^{x^2}y)=x{e}^{x^2}$. Integrate: the right side is $\int x{e}^{x^2}dx$. Why substitution here? Because the derivative of the exponent $x^2$ is $2x$, and we have an $x$ outside — a classic "inner-derivative-is-present" signal. Let $u=x^2$, $du=2xdx$: $\int x{e}^{x^2}dx=\frac{1}{2}\int e^{u}du=\frac{1}{2}{e}^{x^2}+C.$ So $e^{x^2}y=\frac{1}{2}{e}^{x^2}+C$. Divide: $y=\frac{1}{2}+C{e}^{-x^2}$.

Standardize: divide by $x$ (coefficient of $\frac{dy}{dx}$ must be $1$): $\frac{dy}{dx}-\frac{2}{x}y=x^3$. So $P=-\frac{2}{x},Q=x^3$. Mu: $\mu =e^{\int -\frac{2}{x}dx}=e^{-2\ln x}=x^{-2}=\frac{1}{x^2}$. Multiply & collapse: $\frac{d}{dx}\left(x^{-2}y\right)=x^{-2}\cdot x^3=x$. Integrate: $x^{-2}y=\int xdx=\frac{x^2}{2}+C$. Divide: multiply by $x^2$: $y=\frac{x^4}{2}+C{x}^2$.

Standardize: $P=1,Q=x$. Mu: $\mu =e^{\int 1dx}=e^x$. Multiply & collapse: $\frac{d}{dx}(e^{x}y)=x{e}^x$. Integrate by parts: Why parts? We have a product of "polynomial $\times$ exponential" and no chain-rule inner derivative is present, so substitution won't help — parts is the tool that trades a hard product for an easier one. With $u=x,dv=e^{x}dx$: $\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C.$ So $e^{x}y=x{e}^x-e^x+C$. Divide: $y=x-1+C{e}^{-x}$. Apply IC last: $y(0)=2\Rightarrow 2=0-1+C\Rightarrow C=3$. Answer: $y=x-1+3e^{-x}$.

Standardize: divide by $\cos x$ (nonzero on this interval): $\frac{dy}{dx}+\tan xy=\sec x$. So $P=\tan x,Q=\sec x$. Mu: $\mu =e^{\int \tan xdx}=e^{-\ln |\cos x|}=\frac{1}{\cos x}=\sec x$ (since $\cos x>0$ here). Multiply & collapse: $\frac{d}{dx}(\sec xy)=\sec x\cdot \sec x={\sec }^{2}x$. Integrate: $\sec xy=\int {\sec }^{2}xdx=\tan x+C$. Divide: $y=\frac{\tan x+C}{\sec x}=\sin x+C\cos x$ (using $\frac{\tan x}{\sec x}=\sin x$ and $\frac{1}{\sec x}=\cos x$).

This is a Bernoulli equation with $n=2$ — not linear as written (the $x{y}^2$ term multiplies powers). Substitute $v=y^{1-n}=y^{-1}$, so $\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$. Divide the ODE by $y^2$: $y^{-2}\frac{dy}{dx}+y^{-1}=x$. In terms of $v$: $-\frac{dv}{dx}+v=x$, i.e. $\frac{dv}{dx}-v=-x.$ Now it's linear in $v$: $P=-1,Q=-x$. $\mu =e^{\int -1dx}=e^{-x}$. $\frac{d}{dx}(e^{-x}v)=-x{e}^{-x}$. Integrate by parts: $\int -x{e}^{-x}dx=x{e}^{-x}+e^{-x}+C$. So $e^{-x}v=x{e}^{-x}+e^{-x}+C\Rightarrow v=x+1+C{e}^x$. Back-substitute $v=1/y$: $\frac{1}{y}=x+1+C{e}^x$, so $\boxed{y=\frac{1}{x+1+C{e}^x}}$.

$\mu =e^x$ throughout. On $0\le x\le 1$ ($f=1$): $\frac{d}{dx}(e^{x}y)=e^x\Rightarrow e^{x}y=e^x+C_1\Rightarrow y=1+C_1{e}^{-x}$. IC $y(0)=0$: $0=1+C_1\Rightarrow C_1=-1$, so $y=1-e^{-x}(0\le x\le 1).$ At $x=1$: $y(1)=1-e^{-1}$. On $x>1$ ($f=0$): $\frac{d}{dx}(e^{x}y)=0\Rightarrow e^{x}y=C_2\Rightarrow y=C_2{e}^{-x}$. Continuity at $x=1$: $C_2{e}^{-1}=1-e^{-1}\Rightarrow C_2=e-1$, so $y=(e-1)e^{-x}(x>1).$ As $x\to \infty$, $y\to 0$: with the driving turned off, the solution decays.

Rewrite: $\frac{dy}{dx}-\frac{1}{x}y=1$, so $P=-\frac{1}{x},Q=1$. $\mu =e^{\int -\frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$. $\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{1}{x}\cdot 1=\frac{1}{x}$. Integrate: $\frac{y}{x}=\ln x+C$. So $y=x\ln x+Cx$. IC $(1,0)$: $0=1\cdot 0+C\cdot 1\Rightarrow C=0$. Answer: $y=x\ln x$. Shape: at $x=1$, $y=0$; the red curve dips slightly negative just right of $x=1$ (since $\ln x<0$ for $x<1$... on $x>1$ it climbs), touches a minimum where $\frac{dy}{dx}=\ln x+1=0$, i.e. $x=e^{-1}$, then rises steadily — a gently curving line pinned through the marked point.

Case $k=-1$: $\mu =e^x$, $\frac{d}{dx}(e^{x}y)=e^x{e}^{-x}=1$. Integrate: $e^{x}y=x+C$, so $y=(x+C)e^{-x}$. Notice the $x$ out front — the forcing $e^{-x}$ matches the decaying mode $e^{-x}$, so integrating a constant produces a linear factor. Case $k\ne -1$: $\frac{d}{dx}(e^{x}y)=e^{(k+1)x}$, integrate: $e^{x}y=\frac{1}{k+1}{e}^{(k+1)x}+C$, so $y=\frac{1}{k+1}{e}^{kx}+C{e}^{-x}$. No stray $x$ — the forcing and the natural mode differ, so a plain exponential particular solution works. Insight: the $x$-multiplier appears exactly when $Q$'s exponential rate equals the homogeneous rate $-P$. (This is the ODE analogue of resonance.)

Let $\mu =e^{\int P}$, so ${\mu }^{\prime }=\mu P$ (from the derivation's condition). Write $\mu y=\int \mu Qdx+C$. Differentiate both sides with the product rule: $\underset{\frac{d}{dx}(\mu y)}{\underbrace{{\mu }^{\prime }y+\mu y^{\prime }}}=\mu Q.$ Replace ${\mu }^{\prime }=\mu P$: $\mu Py+\mu y^{\prime }=\mu Q$. Divide by $\mu$ (never zero since it's an exponential): $y^{\prime }+Py=Q$. ✓ Exact-equation view: multiplying $y^{\prime }+Py-Q=0$ by $\mu$ makes the left side the exact differential $\frac{d}{dx}(\mu y)$ — the integrating factor is precisely what turns a non-exact form into an exact one.

(a) Standardize (for $x\ne 0$): $\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x}$, $\mu =x$. Then $\frac{d}{dx}(xy)=x\cdot \frac{1}{x}=1$, so $xy=x+C\Rightarrow y=1+\frac{C}{x}$. (b) The standard form has $P=\frac{1}{x}$ and $Q=\frac{1}{x}$, both undefined at $x=0$ — the existence-uniqueness theorem needs $P,Q$ continuous, which fails there. Looking at $y=1+\frac{C}{x}$: as $x\to 0$, the term $\frac{C}{x}$ blows up unless $C=0$. So the only solution that stays finite at $x=0$ is $y=1$, giving $y(0)=1$. Therefore an initial condition $y(0)=y_0$ is admissible only if $y_0=1$; any other value has no solution reaching the origin. (Notice the original equation at $x=0$ reads $0+y=1$, forcing $y(0)=1$ — consistent.)

• L1: Is $\frac{dy}{dx}+y^2=x$ linear? ::: No — $y^2$ is a second power of $y$.
• L2: $\mu$ for $\frac{dy}{dx}+2xy=x$? ::: $e^{x^2}$.
• L3: First move for $\cos x{y}^{\prime }+\sin xy=1$? ::: Divide by $\cos x$ to standardize.
• L4: Substitution for $y^{\prime }+y=x{y}^2$? ::: $v=y^{-1}$ (Bernoulli, $n=2$).
• L5: Which $y(0)$ works for $x{y}^{\prime }+y=1$? ::: Only $y(0)=1$.