Question bank — Second-order linear ODEs — superposition principle, general theory

False — linearity is baare mein hai ki $y$ kaise enter karta hai, $x$ ke baare mein nahi. Yahan $y$ sirf pehli power mein aata hai, isliye yeh linear hai ek variable coefficient $q(x)=x^2$ ke saath.

False — product $y\cdot y^{\prime }$ do cheezein multiply karta hai jo unknown se involve hain, aur linearity yehi forbid karti hai.

True — superposition kehta hai $c_1{y}_1+c_2{y}_2$ kisi bhi constants ke liye ek solution hai, aur $y\equiv 0$ isse solve karta hai, isliye solutions scaling aur addition ke under closed hain. Dekho Linear algebra — vector spaces and bases.

False — zero function isse solve nahi karta ($L[0]=0\ne g$), aur do solutions ko add karne par $L[y_1+y_2]=2g\ne g$ milta hai. Yeh homogeneous space ka ek affine shift hai, vector space nahi.

True — sabse bada derivative second hai, isliye general homogeneous solution mein do free constants hote hain, jo $y(x_0),y^{\prime }(x_0)$ jaisi do conditions se fix hote hain.

True — ek bhi nonzero Wronskian value coefficient matrix ko wahan nonsingular force karti hai, isliye koi bhi nontrivial $(c_1,c_2)$ combination ko zero nahi kar sakta.

True — Abel's theorem deta hai $W(x)=W(x_0)\exp (-\int p)$, aur exponential kabhi zero nahi hota, isliye $W$ all-or-nothing hai.

False — $L[y_1+y_2]=(-2)+(-2)=-4\ne -2$. Free adding/scaling exclusively homogeneous property hai.

True — $L[y_1-y_2]=g-g=0$. Yahi structure theorem $y=y_h+y_p$ ka seed hai.

True — linear ODEs ka existence–uniqueness theorem guarantee karta hai ki solution continuity ke poore interval tak extend hota hai, unlike nonlinear case. Dekho Existence and uniqueness theorems for ODEs.

Dono functions dependent hain ($y_2=3y_1$), isliye $c_1{e}^x+c_2\cdot 3e^x=(c_1+3c_2)e^x$ mein actually ek hi free constant hai — yeh fundamental set nahi hai.

Vanishing Wronskian tabhi conclusive hai jab dono same linear ODE ke solutions hon. Ye do functions genuinely independent hain, bawajood $W\equiv 0$ ke; Wronskian test ka converse us context ke bina fail karta hai.

Tum ek particular solution pure homogeneous family mein add karte ho: $y=c_1{y}_1+c_2{y}_2+y_p$. $y_p$ ko pieces mein distribute karna meaningless hai.

Forcing constant hai, isliye constant $y_p=A$ try karo. Kyunki $L$ ek constant ko constant par map karta hai, $-A=-2\Rightarrow A=2$. Dekho Method of undetermined coefficients.

$L$ functions par linear hai linearity identity $L[c_1{y}_1+c_2{y}_2]=c_{1}L[y_1]+c_{2}L[y_2]$ ke saath; constant $3$ add karne par $L[y]+L[3]=L[y]+3q$ milta hai, $L[y]+3$ nahi. Constants $L$ se fix nahi hote jab tak $q=0$ na ho.

Fundamental-set conclusion ke liye zaroori hai ki dono same homogeneous equation solve karein. Alag-alag ODEs ke independent solutions ek ODE ke solution space ko span karne ke baare mein kuch nahi kehte.

Kyunki $L$ linear hai: $L[c_1{y}_1+c_2{y}_2]=c_{1}L[y_1]+c_{2}L[y_2]$. Jab dono outputs $0$ hote hain toh result kisi bhi scalars ke liye $0$ hota hai; jab dono $g$ hote hain toh result $(c_1+c_2)g$ hota hai, jo $g$ ke barabar sirf tab hota hai jab $c_1+c_2=1$.

Kyunki $e^{rx}$ differentiation ke under khud ko reproduce karta hai, isliye substitute karne par ODE algebraic characteristic equation mein $r$ ke terms mein collapse ho jaata hai. Dekho Characteristic equation — constant coefficient ODEs.

Dependence force karti hai ki $c_1{y}_1+c_2{y}_2=0$ aur iska derivative $c_1{y}_1^{\prime }+c_2{y}_2^{\prime }=0$ dono hon; nontrivial $(c_1,c_2)$ tab exist karta hai jab yeh $2\times 2$ system singular ho, yaani iska determinant $y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ zero ho.

Formula $W(x)=W(x_0)e^{-\int p}$ $W$ ko $x_0$ par uski value ka ek nonzero-exponential multiple banata hai, isliye iska sign aur nonvanishing us single value se fix ho jaate hain — all-or-nothing.

Yeh certify karta hai ki two-parameter family $c_1{y}_1+c_2{y}_2+y_p$ har solution ko capture karti hai; iske bina, exotic solutions family ke bahar chhipe ho sakte hain. Dekho Existence and uniqueness theorems for ODEs.

Monic form theory (Wronskian, Abel, existence) ko ek well-defined $p(x)$ ke saath cleanly padhne laayak banata hai; leading-coefficient ka clutter warna har formula mein ghus jaata.

Har integration ek arbitrary constant introduce karti hai; second order 'do baar integrate' karta hai. Compare karo First-order linear ODEs se, jinmein sirf ek constant hota hai.

Undetermined coefficients sirf special forcings (polynomials, exponentials, sines) ke liye forms guess karta hai; Variation of parameters kisi bhi continuous $g$ ke liye fundamental set se $y_p$ build karta hai.

Haan — $L[0]=0$ hamesha hota hai, yehi wajah hai ki homogeneous solution set ek genuine vector space hai jo zero vector contain karta hai.

Existence–uniqueness ki guarantee sirf us open interval par hold karti hai jahan $p,q,g$ sab continuous hon; discontinuity ke across tum uniqueness ya global extension kho sakte ho, aur $W$ ka all-or-nothing rule us point par toot sakta hai.

Haan, dono ek point par zero hit kar sakte hain; independence sirf require karti hai ki $W\ne 0$ kahin par ho, aur $W(x_1)$ tab bhi nonzero ho sakta hai jab $y_1(x_1)=y_2(x_1)=0$ ho (unke derivatives differ karte hain).

Sirf wahan jahan $g\equiv 0$ ho; agar $g$ kisi bhi sub-interval par nonzero hai toh ODE wahan non-homogeneous hai, aur tumhe us region mein ek particular solution patch karni hogi.

Nahi — tumhe phir bhi do independent solutions chahiye; jab root repeat hota hai, toh doosra $x{e}^{rx}$ hota hai, jo ek two-dimensional solution space preserve karta hai. Dekho Characteristic equation — constant coefficient ODEs.

Nahi — tab woh $y_p$ $y_h$ mein absorb ho jaata hai, isliye tumhe ek genuinely non-homogeneous particular solution dhundh na hogi (aksar guess ko $x$ se multiply karke) taaki $y_h$ aur $y_p$ alag rahen.

Nahi — $y_2$ ko $5y_1$ se replace karna independence collapse kar deta hai; lekin $y_1$ aur $y_1+y_2$ abhi bhi ek valid fundamental set hai, kyunki yeh independent functions ka ek invertible mix hai.