Second-order linear ODEs — superposition principle, general theory

What is a second-order linear ODE?

WHY this form? Dividing through by the leading coefficient $a(x)$ in $a y'' + b y' + c y = h$ gives the "monic" form above. We do this so the theory (existence, Wronskian) reads cleanly.

Worked example Spot the type

Equation	Linear?	Homogeneous?
$y'' + 3y' + 2y = 0$	✅	✅
$y'' + x^2 y = \cos x$	✅	❌ (forcing $\cos x$ )
$y'' + y\,y' = 0$	❌ ( $y\cdot y'$ product)	—
$y'' + \sin(y) = 0$	❌ ( $\sin y$ nonlinear)	—

Why? Linearity is about how $y$ enters, not about $x$ . $x^2$ multiplying $y$ is fine.

The Linear Operator viewpoint (the engine of everything)

Formula Linearity of

$L$ — derived Take any functions $y_1,y_2$ and constants $c_1,c_2$ . Differentiation is linear: $(c_1y_1+c_2y_2)' = c_1y_1' + c_2y_2'$ , and likewise for the second derivative. Substitute:

\begin{aligned} L[c_1y_1+c_2y_2] &= (c_1y_1+c_2y_2)'' + p(c_1y_1+c_2y_2)' + q(c_1y_1+c_2y_2)\\ &= c_1y_1'' + c_2y_2'' + p(c_1y_1'+c_2y_2') + q(c_1y_1+c_2y_2)\\ &= c_1\underbrace{(y_1''+py_1'+qy_1)}_{L[y_1]} + c_2\underbrace{(y_2''+py_2'+qy_2)}_{L[y_2]}\\ &= c_1 L[y_1] + c_2 L[y_2]. \end{aligned}

So $\boxed{L[c_1y_1+c_2y_2] = c_1L[y_1] + c_2L[y_2]}$ . This single identity is the source of the superposition principle.

General solution structure

Figure — Second-order linear ODEs — superposition principle, general theory

Formula Structure theorem —

derived Homogeneous: The general solution of $L[y]=0$ is $y_h = c_1 y_1 + c_2 y_2,$ where $y_1,y_2$ are linearly independent solutions (a fundamental set).

Non-homogeneous: Let $y_p$ be one particular solution of $L[y]=g$ , and let $y$ be any solution. Then $L[y-y_p]=g-g=0$ , so $y-y_p$ is a homogeneous solution $=y_h$ . Rearranging: $\boxed{y = y_h + y_p = c_1y_1 + c_2y_2 + y_p.}$

WHY two constants? A second-order ODE involves a second derivative; integrating "twice" introduces exactly two arbitrary constants. They're pinned down by two initial conditions $y(x_0)=a,\ y'(x_0)=b$ .

Linear independence & the Wronskian

Formula The Wronskian —

derived test Suppose $c_1y_1+c_2y_2=0$ identically. Differentiate once: $c_1y_1'+c_2y_2'=0$ . Stack into a system: $\begin{pmatrix} y_1 & y_2 \\ y_1' & y_2' \end{pmatrix}\begin{pmatrix} c_1 \\ c_2 \end{pmatrix} = \begin{pmatrix}0\\0\end{pmatrix}.$ A nonzero solution $(c_1,c_2)$ exists iff the matrix is singular, i.e. determinant $=0$ . Define the Wronskian: $W(y_1,y_2)(x) = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix} = y_1y_2' - y_2y_1'.$ Test: If $W(x_0)\neq 0$ at some point, then $y_1,y_2$ are linearly independent, and (for solutions of the same ODE) they form a fundamental set.

Worked Example 1 — building & verifying the general solution

Worked example Solve

$y'' - y = 0$ , with $y(0)=2,\ y'(0)=0$ . Step 1. Guess exponential solutions (motivated: derivatives of $e^{rx}$ reproduce $e^{rx}$ ). Sub $y=e^{rx}$ : $r^2e^{rx}-e^{rx}=0\Rightarrow r^2=1\Rightarrow r=\pm1$ . Why? The characteristic equation turns calculus into algebra.

Step 2. Fundamental set $y_1=e^x,\ y_2=e^{-x}$ ; independent (Wronskian $-2$ above). So $y = c_1e^x + c_2e^{-x}.$

Step 3. Apply ICs. $y(0)=c_1+c_2=2$ . $y'=c_1e^x-c_2e^{-x}$ , so $y'(0)=c_1-c_2=0\Rightarrow c_1=c_2$ . Then $c_1=c_2=1$ . Why? Two ICs ⇔ two equations ⇔ pin down two constants uniquely.

Answer: $y = e^x + e^{-x} = 2\cosh x.$ ✅ Check: $y''=2\cosh x = y$ . ✔

Worked Example 2 — non-homogeneous structure

Worked example Solve

$y'' - y = -2$ , with general structure. Step 1 — homogeneous part: from Example 1, $y_h = c_1e^x + c_2e^{-x}$ .

Step 2 — one particular solution: the forcing $-2$ is constant; try a constant $y_p=A$ . Then $y_p''=0$ and $0 - A = -2 \Rightarrow A=2$ . Why guess a constant? The RHS is a constant; a constant input through $L$ stays a constant.

Step 3 — assemble: $y = y_h + y_p = c_1e^x + c_2e^{-x} + 2.$ Why add them? Structure theorem: every solution = (any particular) + (general homogeneous).

Check: $y'' = c_1e^x+c_2e^{-x}$ , so $y''-y = (c_1e^x+c_2e^{-x}) - (c_1e^x+c_2e^{-x}+2) = -2.$ ✔

Existence & Uniqueness (the licence to use this theory)

Common mistakes (steel-manned)

Flashcards

What is the standard (monic) form of a 2nd-order linear ODE?

y'' + p(x)y' + q(x)y = g(x)

When is such an ODE homogeneous?

When the forcing term

g(x)\equiv 0

State the superposition principle.

y_1,y_2

solve

L[y]=0

, then

c_1y_1+c_2y_2

solves

L[y]=0

for all constants

c_1,c_2

Why does superposition fail for non-homogeneous ODEs?

L[c_1y_1+c_2y_2]=(c_1+c_2)g

, which equals

g

only if

c_1+c_2=1

; free scaling isn't allowed.

What is the general solution structure of

L[y]=g

y = y_h + y_p = c_1y_1+c_2y_2+y_p

(general homogeneous + one particular).

Why exactly two arbitrary constants?

Second derivative ⇒ "two integrations" ⇒ two constants, fixed by two initial conditions.

Define the Wronskian of

y_1,y_2

W = y_1y_2' - y_2y_1'

(determinant of

[[y_1,y_2],[y_1',y_2']]

What does

W(x_0)\neq 0

tell you?

y_1,y_2

are linearly independent (and form a fundamental set if both solve the same ODE).

State Abel's theorem.

W(x)=W(x_0)\exp(-\int_{x_0}^x p(t)\,dt)

; so

W

is either identically zero or never zero.

W=0

at a point proof of dependence?

Not in general; only for solutions of the same linear ODE (and even then need it on the whole interval).

State the existence–uniqueness condition.

p,q,g

continuous on interval

I\ni x_0

, the IVP has a unique solution on all of

I

Key property of the operator

L[y]=y''+py'+qy

It is linear:

L[c_1y_1+c_2y_2]=c_1L[y_1]+c_2L[y_2]

Recall Feynman: explain to a 12-year-old

Imagine a swing. There are only a couple of "basic" ways it can swing on its own (these are $y_1$ and $y_2$ ). Any other swinging you can get by mixing these two basics in different amounts — push twice as hard on one, half as hard on the other. That mixing is superposition. Now if someone keeps pushing the swing (that's the forcing $g$ ), you first find one way it settles ( $y_p$ ), then add the free basic swings on top. The Wronskian is just a quick check that your two basic swings are really different and not secretly the same swing in disguise.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, second-order linear ODE matlab equation jisme $y''$ , $y'$ aur $y$ aate hain, lekin sirf "pehli power" mein — koi $y^2$ nahi, koi $\sin y$ nahi. Standard form hai $y'' + p(x)y' + q(x)y = g(x)$ . Agar $g(x)=0$ ho to bolते hain homogeneous, warna non-homogeneous (kyunki ek forcing term push kar raha hai).

Sabse important cheez hai superposition principle. Agar $y_1$ aur $y_2$ dono homogeneous equation ko solve karte hain, to inka koi bhi mix $c_1y_1+c_2y_2$ bhi solution hoga. Yeh kaam karta hai kyunki operator $L[y]=y''+py'+qy$ linear hai — derivative linear hota hai, isliye $L$ bhi. Lekin yaad rakho: yeh free mixing sirf homogeneous case mein chalta hai. Non-homogeneous mein agar tum do solutions add karoge to forcing double ho jaata hai, isliye wahan formula alag hai: $y = y_h + y_p$ , yaani general homogeneous solution plus koi ek particular solution.

Do constants kyun? Kyunki second derivative hai, "do baar integrate" karna padta hai, to do arbitrary constants aate hain, jinko hum do initial conditions $y(x_0)$ aur $y'(x_0)$ se fix karte hain. Ab yeh check karne ke liye ki tumhare $y_1,y_2$ sach mein alag hain (ek doosre ke multiple nahi), hum Wronskian $W=y_1y_2'-y_2y_1'$ nikaalte hain. Agar kisi ek point pe $W\neq0$ , to woh independent hain — aur Abel ke theorem se $W$ ya to hamesha zero hota hai ya kabhi nahi, isliye ek hi point check karna kaafi hai.

Practical recipe yaad rakho: pehle homogeneous solve karo ( $y_h$ , do constants), phir ek particular solution dhundo ( $y_p$ , sirf ek baar), phir add kar do. Spring-mass, RLC circuit, sab isi theory pe chalte hain — isliye yeh foundation hai.