Question bank — Integrating factors for non-exact equations

True in principle — ek integrating factor smooth $Mdx+Ndy$ ke liye locally hamesha exist karta hai, lekin ye dono $x$ aur $y$ par depend kar sakta hai aur dhundna utna hi mushkil ho sakta hai jitna ODE directly solve karna.

True — constant function $1$ ek bilkul sahi $\mu$ hai; ek exact equation sirf wo special case hai jahan koi rescaling ki zaroorat nahi.

False — $\mu$ nonzero hona chahiye; $0$ se multiply karne par sab kuch $0=0$ ban jaata hai aur solution curves bilkul destroy ho jaati hain.

False — kyunki $\mu \ne 0$ hai, equation $\mu Mdx+\mu Ndy=0$ bilkul usi set par vanish hoti hai jahan pehle hoti thi, isliye solution curves identical hain.

True — aisa hota hai (jaise ki ye $\mu (xy)$ bhi admit kar sakta hai), aur dono ise exact banate hain; resulting potentials alag hote hain lekin same solution family describe karte hain.

True — wo ratio ${\mu }^{\prime }/\mu$ ke barabar hai, jo $x$ ka pure function hai, aur integrate hokar genuine $\mu (x)=\exp \int gdx$ deta hai.

False — iske solutions ab bhi hain; bas aap do aasaan single-variable formulas use nahi kar sakte aur doosre ansätze jaise $\mu (xy)$ ya $\mu (x+y)$ try karne padte hain.

False — ye sirf tab exact hoti hai jab ratio sach mein single-variable tha; agar aapne invalid ratio par formula force kiya, toh $M_y^{\ast }\ne N_x^{\ast }$ ab bhi rahega.

Error ye hai ki $x$-test ke baad ruk gaye; aapko $y$-test $\frac{N_x-M_y}{M}=\frac{1}{y}$ bhi try karna chahiye tha, jo pure $y$ hai aur $\mu =y$ deta hai.

Numerator ka sign galat hai — $y$-case mein $\frac{N_x-M_y}{M}$ hota hai; minus isliye aata hai kyunki $-M{\mu }^{\prime }=\mu (M_y-N_x)$ side change karta hai.

Aapko puri equation, yaani dono $M$ aur $N$, ko $\mu$ se multiply karna hoga; ek term ko scale karne se equation ka meaning toot jaata hai aur wo exact nahi hogi.

Integration ka 'constant' $\varphi (y)$ hai, na ki koi true constant $C$; aapko $y$ mein differentiate karke $N^{\ast }$ se match karna hoga, tabhi $F=C$ likhne se pehle wo pin down hoga.

$\mu =\exp (\int \frac{1}{x}dx)=\exp (\ln x)=x$ hai, na ki $\frac{1}{x}$; exponential bhool jaana (aur exponentiate karne ki jagah sirf integrate karna) classic slip hai.

Formula exactness guarantee karta hai sirf tab jab ratio genuinely $x$-only tha; ek re-check $M_y^{\ast }=N_x^{\ast }$ un arithmetic errors ko pakad leta hai jo quietly ratio ko poison kar deti hain.

Kyunki ${\mu }_y=0$ hone se $M{\mu }_y$ term khatam ho jaata hai, sirf $N{\mu }^{\prime }=\mu (M_y-N_x)$ bachta hai, jo $x$ mein ek separable ODE hai, na ki two-variable PDE.

Left side ${\mu }^{\prime }/\mu$ sirf $x$ ka function hai; ye right side ke barabar tab hi ho sakta hai jab right side mein bhi koi $y$ na ho, warna equation self-contradictory ban jaati hai.

Exactness ka matlab hai $Mdx+Ndy=dF$, toh ODE $dF=0$ ban jaata hai; jis quantity ka zero total differential ho, wo solutions ke along constant rehta hai, isliye $F=C$.

$y^{\prime }+Py=Q$ ko $(Py-Q)dx+dy=0$ likhne par $\frac{M_y-N_x}{N}=P(x)$ milta hai, jo pure $x$ hai, isliye general formula exactly linear factor reproduce karta hai.

Kyunki $M=F_x,N=F_y$ hone se $M_y=F_{xy}$ aur $N_x=F_{yx}$ milta hai, jo tab hi equal hote hain jab smooth $F$ ke liye mixed partials commute karein.

$Mdx+Ndy=dF$ ka matlab hai ki $(M,N)$, $F$ ka gradient hai, isliye ye ek conservative field hai jiska potential $F$ aur curl $M_y-N_x=0$ hai.

Kyunki general $\mu (x,y)$ ek aisi PDE satisfy karta hai jo original problem jitni hi mushkil hai; ek variable tak restrict karne se ye ek solvable ODE ban jaata hai, aur yehi poora practical fayda hai.

Nahi — $M_y=N_x$ ka matlab hai ye pehle se exact hai, toh $\mu =1$; surface complexity ki parwah kiye bina seedha $F$ dhundhne jao.

Nahi — $\frac{M_y-N_x}{N}$ mein $N$ se divide hota hai, isliye $N\equiv 0$ ise undefined bana deta hai; lekin $N=0$ ka matlab hai $Mdx=0$, jo trivially separable hai.

Theek hai — constant ek valid single-variable function hai, jo $\mu =e^{cx}$ ya $e^{cy}$ deta hai; constants sabse aasaan allowed case hain, koi failure nahi.

Iska matlab hai $M_y=N_x$, isliye $g(x)=0$ aur $\mu =\exp (0)=1$; formula gracefully "already exact" return karta hai, jo consistency confirm karta hai.

Haan — $x=0$ par factor vanish karta hai, isliye exactness (aur potential ki validity) us line par toot sakti hai; $x=0$ cross karne wali solutions ko alag treatment ki zaroorat ho sakti hai kyunki $\mu \ne 0$ assume kiya gaya tha.

Haan — methods exclusive nahi hain; agar equation separable hai toh wo kisi $\mu$ ke baad exact bhi ho sakti hai, aur dono raaste same solution family tak pahunchte hain.

Formula $\frac{N_x-M_y}{M}$ mein $M=0$ se divide hota hai aur ye undefined ho jaata hai; lekin $M=0$ hone par $Ndy=0$ milta hai, jo phir se ek degenerate/separable case hai jise directly handle kiya jaata hai.