Second-order linear ODEs — superposition principle, general theory: Exercises

4.6.9 · D4 · Maths › Ordinary Differential Equations › Second-order linear ODEs — superposition principle, general

Shuru karne se pehle, chaar words jinka hum baar baar use karenge, har ek ek picture se juda hua:

Definition Chaar anchors

Ek operator $L$ ek machine hai jo ek function $y (x)$ leta hai aur ek naya function return karta hai. Yahan $L [y] = y^{''} + p (x) y^{'} + q (x) y$ . Ek box ki tarah socho: function left se andar jaata hai, function right se bahar aata hai.
Homogeneous ka matlab hai right-hand side zero ho: $L [y] = 0$ . Koi baahri "push" nahi.
Forcing term $g (x)$ woh hai jo $L [y] = g (x)$ ke right-hand side par baitha hai — system par lagaaya gaya baahri "push" (jaise driving voltage ya applied force). Jab $g \equiv 0$ ho toh hum homogeneous hain; jab $g \neq = 0$ ho toh hum non-homogeneous hain.
Do functions ka Wronskian yeh number hai $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ . Yeh ek detector hai: kisi point par nonzero $\Rightarrow$ do functions genuinely alag directions hain, scaled copies nahi. Dekho Abel's theorem iske all-or-nothing hone ki wajah ke liye.

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

Upar wale four-anchor figure ko ek control panel ki tarah left se right padhо. Sabse baayein cyan box operator $L$ hai: ek function $y$ uske left wire se enter karta hai aur transformed function $L [y]$ right se bahar nikalti hai — yeh anchor 1 ki "machine" hai. Upper branch homogeneous case dikhata hai, jahan output wire par $L [y] = 0$ likha hai (koi bhi cheez system ko push nahi karti). Lower branch forcing term $g (x)$ ko ek amber input arrow ke roop mein aate dikhata hai aur output par $L [y] = g$ ke roop mein appear karta hai — anchor 3 ka baahri push. Right panel Wronskian ko ek chhota $2 \times 2$ determinant ki tarah draw karta hai jisme $y_{1}, y_{2}$ apni derivatives $y_{1}^{'}, y_{2}^{'}$ ke upar stacked hain: amber "area" jo yeh measure karta hai woh nonzero hota hai bilkul tabhi jab do solution-arrows genuinely alag directions mein hों (anchor 4). Yeh panel dhyaan mein rakho — neeche ke har exercise mein se ek in chaar boxes ko poke karta hai.

Intuition Ek theory fact jo humein L5 par chahiye — repeated roots

Ek constant-coefficient homogeneous ODE $a y^{''} + b y^{'} + cy = 0$ ke liye, trick yeh hai ki $y = e^{r x}$ try karo, calculus ko algebraic characteristic equation $a r^{2} + b r + c = 0$ mein badal do (dekho Characteristic equation — constant coefficient ODEs).

Distinct real roots $r_{1} \neq = r_{2}$ : fundamental set ${e^{r_{1} x}, e^{r_{2} x}}$ — do clearly alag directions.
Repeated root $r_{1} = r_{2} = r$ : exponential sirf ek solution $e^{r x}$ deta hai. Doosra independent solution $x e^{r x}$ hai — $x$ ka ek extra factor. Fundamental set ${e^{r x}, x e^{r x}}$ .

WHY extra $x$ : jab do roots collide karte hain toh sirf ek exponential direction bachta hai, isliye solution space ek-dimensional lagti hai — lekin ek second-order ODE ka solution space hamesha two-dimensional hota hai. Jo direction missing hai woh $x e^{r x}$ ke roop mein dobara appear hoti hai. L5.3 exactly yahi case hai apne sabse clean form mein ( $r = 0$ ).

Level 1 — Recognition

Kya tum type naam le sakte ho aur pieces padh sakte ho?

L1.1

Har equation ko classify karo (a) linear ya nonlinear, (b) homogeneous ya non-homogeneous. Jahan linear ho wahan coefficient functions $p (x), q (x)$ aur forcing $g (x)$ do. $(i) y^{''} + 4 y^{'} + 3 y = 0 (ii) y^{''} + x^{2} y = e^{x} (iii) y^{''} + y y^{'} = sin x$

Recall Solution

WHAT hum karte hain: har ek ko standard form $y^{''} + p (x) y^{'} + q (x) y = g (x)$ se compare karo. WHY yeh ek saath sab kuch answer karta hai: standard form ek template with named slots hai — jis moment ek equation isse match karti hai, "linear?" ka jawab mil jaata hai (kya $y$ sirf un slots ke through enter karta hai, first power mein, koi products/nonlinear functions nahi?), aur "homogeneous?" ka jawab right-hand slot $g (x)$ padhne se milta hai (zero hai ya nahi). Ek comparison teeno cousin questions settle kar deta hai.

(i) Linear (har ek $y, y^{'}, y^{''}$ ek baar, first power mein appear karta hai). Homogeneous ( $g = 0$ ). Padho $p = 4$ , $q = 3$ , $g = 0$ .

(ii) Linear — $x^{2}$ , $y$ ko multiply karta hai, aur $x$ input variable hai, $y$ nahi; linearity sirf is baat ki parwah karti hai ki $y$ kaise enter karta hai. Non-homogeneous, $g = e^{x}$ . Yahan $p = 0$ , $q = x^{2}$ , $g = e^{x}$ .

(iii) Nonlinear — term $y y^{'}$ unknown aur uski derivative ka product hai, jo template ke kisi slot mein allowed nahi hai. Koi $p, q, g$ apply nahi hota.

L1.2

Neeche diye gaye pairs mein se kaunsa kisi second-order homogeneous linear ODE ka fundamental set ban sakta hai (matlab kya woh linearly independent hain)? Sirf yes/no aur ek-line reason do. $(a) {e^{x}, e^{2 x}} (b) {e^{x}, 5 e^{x}} (c) {sin x, cos x}$

Recall Solution

WHAT hum karte hain: poochhte hain "kya ek doosre ka constant multiple hai?" WHY yeh sahi test hai: ek fundamental set ko two-dimensional solution space span karni chahiye; agar ek function doosre ka sirf ek scaled copy hai, toh dono milke ek single direction mein point karte hain aur sirf ek one-parameter family produce kar sakte hain — do initial conditions meet karne ke liye yeh kafi nahi. Toh "constant multiple?" directly fundamental-set eligibility decide karta hai.

(a) Haan, independent — $e^{2 x} / e^{x} = e^{x}$ constant nahi hai.

(b) Nahi — $5 e^{x}$ $e^{x}$ ka constant multiple hai. Woh sirf ek direction span karte hain, ek parameter.

(c) Haan, independent — $tan x$ constant nahi hai, isliye $sin x$ , $cos x$ ka fixed multiple nahi hai.

Level 2 — Application

Standard machinery chalao.

L2.1

Verify karo ki $y_{1} = e^{- x}$ aur $y_{2} = e^{- 3 x}$ dono $y^{''} + 4 y^{'} + 3 y = 0$ solve karte hain, phir general solution likho.

Recall Solution

WHAT & WHY — har candidate substitute karo. Hum plug in karte hain kyunki "solution hai" ka matlab hai substitution ke baad equation hold kare.

$y_{1} = e^{- x}$ ke liye: $y_{1}^{'} = - e^{- x}$ , $y_{1}^{''} = e^{- x}$ . Tab $e^{- x} + 4 (- e^{- x}) + 3 e^{- x} = (1 - 4 + 3) e^{- x} = 0. ✓$ $y_{2} = e^{- 3 x}$ ke liye: $y_{2}^{'} = - 3 e^{- 3 x}$ , $y_{2}^{''} = 9 e^{- 3 x}$ . Tab $9 e^{- 3 x} + 4 (- 3 e^{- 3 x}) + 3 e^{- 3 x} = (9 - 12 + 3) e^{- 3 x} = 0. ✓$ Woh independent hain (ratio $e^{- 3 x} / e^{- x} = e^{- 2 x}$ constant nahi), isliye $y = c_{1} e^{- x} + c_{2} e^{- 3 x} .$

L2.2

$y_{1} = e^{- x}, y_{2} = e^{- 3 x}$ ke liye Wronskian $W (y_{1}, y_{2})$ compute karo aur confirm karo ki yeh kabhi zero nahi hota.

Recall Solution

WHAT hum karte hain: $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ . WHY: yeh determinant exactly woh condition hai jab "koi nonzero $(c_{1}, c_{2})$ dono $y$ aur $y^{'}$ ko simultaneously zero nahi kar sakta" — matlab independence. $W = e^{- x} (- 3 e^{- 3 x}) - e^{- 3 x} (- e^{- x}) = - 3 e^{- 4 x} + e^{- 4 x} = - 2 e^{- 4 x} .$ Kyunki har $x$ ke liye $e^{- 4 x} > 0$ , $W = - 2 e^{- 4 x} \neq = 0$ everywhere. Independent — L2.1 confirm karta hai.

L2.3

IVP solve karo $y^{''} + 4 y^{'} + 3 y = 0$ , $y (0) = 2$ , $y^{'} (0) = - 4$ .

Recall Solution

Step 1. General solution (L2.1 se): $y = c_{1} e^{- x} + c_{2} e^{- 3 x}$ .

Step 2 — conditions apply karo. WHY: ek second-order ODE "do baar" integrate hoti hai, jo do constants produce karta hai; do initial conditions unhe uniquely pin down karti hain (existence–uniqueness se guarantee). $y (0) = c_{1} + c_{2} = 2.$ $y^{'} = - c_{1} e^{- x} - 3 c_{2} e^{- 3 x}$ , isliye $y^{'} (0) = - c_{1} - 3 c_{2} = - 4$ .

Step 3 — pair solve karo. $c_{1} + c_{2} = 2$ ko $- c_{1} - 3 c_{2} = - 4$ mein add karo: $- 2 c_{2} = - 2 \Rightarrow c_{2} = 1$ , hence $c_{1} = 1$ . $y = e^{- x} + e^{- 3 x} .$ Check: $y (0) = 1 + 1 = 2$ ✓; $y^{'} (0) = - 1 - 3 = - 4$ ✓.

Level 3 — Analysis

Structure ke baare mein reason karo, sirf handle mat cranko.

L3.1

Abel's theorem use karke, kisi bhi do solutions ka Wronskian dhundho $y^{''} + \frac{1}{x} y^{'} + q (x) y = 0 (x > 0)$ ek constant tak, bina solutions jaane. Phir explain karo ki $W$ ek single isolated point par zero kyun nahi ho sakta.

Recall Solution

WHAT & WHY. Abel's theorem kehta hai $W^{'} = - p (x) W$ , isliye $W (x) = W (x_{0}) exp (- \int p)$ . Yeh hume $p$ se akele $W$ ki shape jaanne deta hai, koi solving ki zaroorat nahi.

Yahan $p (x) = 1/ x$ . Isliye $W (x) = C exp (- \int \frac{1}{x} d x) = C e^{- l n x} = \frac{C}{x} .$ $x > 0$ ke liye: agar $C = 0$ toh $W \equiv 0$ ; agar $C \neq = 0$ toh $W = C / x \neq = 0$ har $x > 0$ ke liye. Koi tarika nahi ki $W$ ek isolated point par zero ho aur baaki jagah nonzero — exponential kabhi zero nahi hoti, isliye $W$ all-or-nothing hai. Isliye ek convenient point check karna kaafi hai.

L3.2

Do functions $f (x) = x^{2}$ aur $g (x) = x ∣ x ∣$ ko $I = (- 1, 1)$ par diya gaya hai. Dikhao ki $W (f, g) \equiv 0$ everywhere hai, phir bhi $f, g$ linearly independent hain. Yeh Wronskian test ka CONTRADICTION kyun nahi hai?

Recall Solution

Piecewise set up karo. $x \geq 0$ ke liye, $g = x^{2}$ ; $x < 0$ ke liye, $g = - x^{2}$ .

Region $x \geq 0$ : $f = x^{2}, g = x^{2}$ , isliye $g = f$ wahan aur $W = f g^{'} - g f^{'} = x^{2} (2 x) - x^{2} (2 x) = 0$ .

Region $x < 0$ : $f = x^{2}, g = - x^{2}$ , isliye $g^{'} = - 2 x$ . Tab $W = x^{2} (- 2 x) - (- x^{2}) (2 x) = - 2 x^{3} + 2 x^{3} = 0$ .

Isliye $W \equiv 0$ pore $I$ par.

Independence. Maano $c_{1} x^{2} + c_{2} x ∣ x ∣ = 0$ sab $x \in I$ ke liye. "Sab $x$ ke liye" ka matlab hai equation har us point par hold karni chahiye jo hum plug in karna chaahein — isliye hum convenient points choose karne ke liye free hain, aur har choice $c_{1}, c_{2}$ ke baare mein ek sach equation deti hai. Hume sirf itni equations chahiye jo $c_{1} = c_{2} = 0$ force karein; yahan do well-chosen points kaafi hain kyunki woh do unknowns mein do independent linear equations dete hain.

$x = 1$ par (jahan $x ∣ x ∣ = 1$ hai): $c_{1} + c_{2} = 0$ .
$x = - 1$ par (jahan $x ∣ x ∣ = - 1$ hai): $c_{1} - c_{2} = 0$ .

Add karne par $2 c_{1} = 0 \Rightarrow c_{1} = 0$ , phir $c_{2} = 0$ . Sirf wahi constants jo kaam karte hain woh dono zero hain, isliye $f, g$ linearly independent hain. (Hume baaki jagah identity verify karne ki zaroorat nahi thi; yeh dikhana ki sirf possible constants zero hain already independence settle kar deta hai.)

Contradiction kyun nahi: clean " $W \equiv 0 \Rightarrow$ dependent" converse sirf tab hold karta hai jab dono functions ek hi linear ODE with continuous coefficients ke solutions hon. Yahan $f, g$ aise nahi hain — $g = x ∣ x ∣$ $x = 0$ par ek way mein twice-differentiable nahi hai jo nice $p, q$ ke consistent ho. Isliye Wronskian test generally one-directional rehta hai: $W \neq = 0$ ek point par $\Rightarrow$ independent, hamesha; reverse ke liye ODE context chahiye (Abel's theorem ke zariye).

L3.3

Maano $y_{p}$ , $L [y] = g$ ka ek solution hai aur $\tilde{y}_{p}$ doosra. Prove karo ki $y_{p} - \tilde{y}_{p}$ homogeneous equation solve karta hai, aur is se argue karo ki general solution ki form $y_{h} + y_{p}$ hoti hai.

Recall Solution

WHAT hum karte hain — $L$ ko difference par apply karo. Kyunki $L$ linear hai (superposition): $L [y_{p} - \tilde{y}_{p}] = L [y_{p}] - L [\tilde{y}_{p}] = g - g = 0.$ Isliye koi bhi do particular solutions ek homogeneous solution se differ karte hain.

General form. Maano $y$ , $L [y] = g$ ka koi bhi solution hai aur ek particular $y_{p}$ fix karo. Tab $L [y - y_{p}] = g - g = 0$ , isliye $y - y_{p}$ koi homogeneous solution $y_{h}$ hai. Rearrange karne par, $y = y_{h} + y_{p} = c_{1} y_{1} + c_{2} y_{2} + y_{p} .$ Kyunki $y$ arbitrary tha, har solution is shape ka hai — kuch miss nahi hota.

Level 4 — Synthesis

Ek poora solution banane ke liye kai tools combine karo.

L4.1

Completely solve karo: $y^{''} - y = - 2$ , $y (0) = 3$ , $y^{'} (0) = 1$ . (Homogeneous part known hai: $y_{h} = c_{1} e^{x} + c_{2} e^{- x}$ .)

Recall Solution

Step 1 — particular solution. Forcing $- 2$ ek constant hai. WHY constant guess karo? Ek constant $A$ ko $L$ mein feed karo: $A^{''} = 0$ , isliye $L [A] = - A$ ; $- A = - 2$ match karne par $A = 2$ milta hai. Try karo $y_{p} = 2$ : indeed $0 - 2 = - 2$ ✓. (Yeh Method of undetermined coefficients ka sabse simple case hai.)

Step 2 — assemble. Structure theorem se, $y = y_{h} + y_{p} = c_{1} e^{x} + c_{2} e^{- x} + 2.$

Step 3 — conditions apply karo. $y (0) = c_{1} + c_{2} + 2 = 3 \Rightarrow c_{1} + c_{2} = 1.$ $y^{'} = c_{1} e^{x} - c_{2} e^{- x}$ , isliye $y^{'} (0) = c_{1} - c_{2} = 1$ . Add karo: $2 c_{1} = 2 \Rightarrow c_{1} = 1$ , phir $c_{2} = 0$ . $y = e^{x} + 2.$ Check: $y^{''} = e^{x}$ , $y^{''} - y = e^{x} - (e^{x} + 2) = - 2$ ✓; $y (0) = 1 + 2 = 3$ ✓; $y^{'} (0) = 1$ ✓.

L4.2

Solve karo $y^{''} + 4 y^{'} + 3 y = 9 x$ (sirf general solution). Level 2 se homogeneous set ${e^{- x}, e^{- 3 x}}$ use karo.

Recall Solution

Step 1 — particular solution. Forcing $9 x$ ek degree-1 polynomial hai, isliye try karo $y_{p} = a x + b$ (ek general degree-1 polynomial). WHY yeh form: ek linear polynomial ki derivatives degree $\leq 1$ ki polynomial rehti hain, isliye equation term-by-term balance ho sakti hai. $y_{p}^{'} = a, y_{p}^{''} = 0.$ $y_{p}^{''} + 4 y_{p}^{'} + 3 y_{p} = 0 + 4 a + 3 (a x + b) = 3 a x + (4 a + 3 b) .$ $9 x + 0$ se match karo: $x$ ka coefficient $3 a = 9 \Rightarrow a = 3$ deta hai; constant $4 a + 3 b = 0 \Rightarrow 12 + 3 b = 0 \Rightarrow b = - 4$ deta hai. Isliye $y_{p} = 3 x - 4$ .

Step 2 — assemble. $y = c_{1} e^{- x} + c_{2} e^{- 3 x} + 3 x - 4.$ Particular part check karo: $y_{p}^{''} = 0, y_{p}^{'} = 3$ , $0 + 12 + 3 (3 x - 4) = 12 + 9 x - 12 = 9 x$ ✓.

Level 5 — Mastery

Prove karo, generalise karo, ya degenerate case handle karo.

L5.1

Prove karo ki superposition principle non-homogeneous equations ke liye free scaling allow nahi karta, aur exactly woh ek linear combination of do particular solutions state karo jo solution HAI.

Recall Solution

Maano $L [y_{1}] = g$ aur $L [y_{2}] = g$ . Linearity se, $L [c_{1} y_{1} + c_{2} y_{2}] = c_{1} L [y_{1}] + c_{2} L [y_{2}] = (c_{1} + c_{2}) g .$ Yeh $g$ ke barabar hoga iff $c_{1} + c_{2} = 1$ . Isliye free scaling allowed nahi hai: sirf woh combinations jिनके coefficients ka sum $1$ ho (affine combinations) solutions rehte hain. Specifically $c_{1} = 1, c_{2} = - 1$ deta hai $L [y_{1} - y_{2}] = 0$ — difference homogeneous hai, jo structure theorem का seed hai. Free scaling exclusively ek homogeneous property hai, jahan $g = 0$ kisi bhi $c_{1}, c_{2}$ ke liye $(c_{1} + c_{2}) \cdot 0 = 0$ bana deta hai.

L5.2

Consider karo $y^{''} - y = 0$ sab $R$ par. Tumhe bataya gaya hai ki $y_{1} = cosh x$ aur $y_{2} = sinh x$ solutions hain. Dikhao ki woh fundamental set banate hain, aur general solution ko exponential basis mein re-express karo confirm karne ke liye ki yeh same solution space hai.

Recall Solution

Solutions? $(cosh x)^{''} = cosh x = y_{1}$ ✓; $(sinh x)^{''} = sinh x = y_{2}$ ✓.

Wronskian se independence. $y_{1}^{'} = sinh x, y_{2}^{'} = cosh x$ : $W = cosh x cosh x - sinh x sinh x = cosh^{2} x - sinh^{2} x = 1 \neq = 0.$ Everywhere nonzero ⇒ fundamental set ⇒ general solution $y = A cosh x + B sinh x$ .

Same space as ${e^{x}, e^{- x}}$ . $cosh x = \frac{e ^{x} + e ^{- x}}{2}, sinh x = \frac{e ^{x} - e ^{- x}}{2}$ use karke, $A cosh x + B sinh x = \frac{A + B}{2} e^{x} + \frac{A - B}{2} e^{- x} = c_{1} e^{x} + c_{2} e^{- x},$ jahan $c_{1} = \frac{A + B}{2}, c_{2} = \frac{A - B}{2}$ . Koi bhi $(A, B)$ kisi $(c_{1}, c_{2})$ par map hota hai aur vice versa — dono "bases" identical two-dimensional solution space span karte hain. Yeh exactly ek change of basis hai: same plane, different arrows.

L5.3 (degenerate / limiting case)

Directly solve karo $y^{''} = 0$ (koi characteristic-equation shortcut nahi). Fundamental set kya hai, $W$ kya hai, aur yeh "boundary case" kyun hai jahan characteristic equation ki do roots collide karti hain?

Recall Solution

Direct integration. Equation kehti hai second derivative everywhere zero hai. Ek baar integrate karo: $y^{'} = c_{1}$ (ek constant, kyunki uski derivative $y^{''}$ $0$ hai). Dobara integrate karo: $y = c_{1} x + c_{2}$ . Isliye fundamental set hai ${y_{1} = 1, y_{2} = x}, general solution y = c_{1} x + c_{2} .$ (Note karo do arbitrary constants $c_{1}, c_{2}$ — exactly woh jo ek second-order equation produce karni chahiye.)

Wronskian. $y_{1} = 1, y_{2} = x$ ke saath $y_{1}^{'} = 0, y_{2}^{'} = 1$ hai, isliye $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'} = (1) (1) - (x) (0) = 1 \neq = 0.$ Everywhere nonzero ⇒ ${1, x}$ genuinely independent hai ⇒ valid fundamental set.

Boundary case kyun. $y = e^{r x}$ likhne par $y^{''} = 0$ characteristic equation $r^{2} = 0$ ban jaati hai, ek repeated root $r = 0$ . Jab do roots coincide karte hain, exponential family sirf ek independent solution deti hai, $e^{0 x} = 1$ ; doosre mein repeated-root rule se $x$ ka extra factor hona chahiye, matlab $x e^{0 x} = x$ . Yeh exactly wahi ${1, x}$ hai jo humne haath se dhundha — koi coincidence nahi. Isliye $y^{''} = 0$ do distinct exponential roots ke ek mein merge hone ki sabse clean picture hai, aur exactly isliye Characteristic equation — constant coefficient ODEs mein repeated-root formula mein $x$ ka woh factor hota hai.