Question bank — Separable ODEs — technique, implicit solutions

False — $\frac{dy}{dx}=x+y$ ke right side par ek sum hai jo kabhi pure-$x$ times pure-$y$ product mein factor nahi hota; uske liye integrating factors chahiye.

True — exponent mein sum split ho jaata hai: $e^{x+y}=e^x{e}^y$, toh $g(x)=e^x$, $h(y)=e^y$; ek sum ka exponential ek product hota hai.

False — yeh ek fully valid solution hai; agar $y$ ko isolate karne mein $\pm$ quadratic formula chahiye toh implicit form par rukna aksar cleaner aur zyada honest hota hai.

True — dono sides par ek-ek integration se do constants milte hain, lekin woh merge hokar ek single $C$ ban jaate hain; ODE ki order (yahan 1) count fix karti hai.

False — $dy/dx$ ek limit hai, fraction nahi; yeh split ek valid shorthand hai substitution/chain-rule step ke liye: $\int \frac{1}{h(y)}\frac{dy}{dx}dx=\int \frac{dy}{h(y)}$.

False — agar $h(y_0)=0$ hai, toh constant $y\equiv y_0$ ODE ko solve karta hai lekin $h(y)$ se division usse mita deta hai; woh equilibrium solutions haath se restore karne padte hain.

False — yeh $\ln |y|+C$ hai; modulus exactly wahi cheez hai jo baad mein constant ko $\pm$ sign absorb karne deta hai jab aap exponentiate karte ho.

True — yahan $h(y)=y$ aur $h(0)=0$ hai, lekin $y\equiv 0$ ko $A=0$ rakhkar recover kiya ja sakta hai, toh is case mein kuch truly lost nahi hai (logistic ke unlike).

True — yeh $g(x)h(y)$ hai jahan $g=\sin x$ aur $h=1/y^2$; ek quotient jisme upar pure-$x$ aur neeche pure-$y$ ho, woh bhi factor karta hai.

False — yeh double-count karta hai; do arbitrary constants ka difference ek arbitrary constant hota hai, toh ek side par single $C$ complete hai.

Error hai bahut jaldi rukna: $y(1-y)$ se divide karne par constants $y\equiv 0$ aur $y\equiv 1$ delete ho gaye, jo genuine solutions hain aur family ke saath report karne chahiye.

Partial fractions mein sign error hai: sahi split hai $\frac{1}{y}+\frac{1}{1-y}$ (dono plus), kyunki $\frac{1}{y}+\frac{1}{1-y}=\frac{(1-y)+y}{y(1-y)}=\frac{1}{y(1-y)}$.

Inner derivative se minus sign missing hai: $\frac{d}{dy}(1-y)=-1$, toh $\int \frac{dy}{1-y}=-\ln |1-y|+C$.

Illegal move — aap sum $x+y$ se $x$ ko pull off nahi kar sakte; sirf ek product factor hota hai, aur $x+y$ product nahi hai.

Ek sum ko exponentiate karne par woh ek product banta hai, sum nahi: $y=e^{x^3/3}\cdot e^C=A{e}^{x^3/3}$, na ki $e^{x^3/3}+e^C$.

Implicit solutions bhi constant carry karte hain: yeh $y+y^2=x^2+C$ hai, aur $C$ wahi hai jo ek initial condition pin down karti hai.

Ek $C$ kaafi hai; dono sides par likhna redundant hai kyunki do constants merge hokar ek ban jaate hain.

Kyunki separation har $y$ ko left aur har $x$ ko right bhejta hai; yeh clean split tabhi possible hai jab do variables multiply (ya divide) ho rahe hon, kabhi add nahi.

Yahan yeh ek specific chain-rule substitution ka faithful shorthand hai jo provably correct hai; kahin aur $dy$ aur $dx$ ko blindly manipulate karne ki aisi koi guarantee nahi hoti.

$1/y$ ka antiderivative negative $y$ ke liye bhi valid hona chahiye; modulus formula ko $y=0$ ke dono sides par sahi rakhta hai, aur uska sign baad mein arbitrary constant mein absorb ho jaata hai.

Agar $h(y_0)=0$ hai toh constant $y\equiv y_0$ se $y^{\prime }=0=g(x)\cdot h(y_0)$ milta hai, toh yeh ODE satisfy karta hai — lekin yahi woh value hai jisse hum divide kar rahe the, toh algebra ne usse chupa diya.

$y$ ke liye solve karne par quadratic $\pm$ branch ya transcendental tangle aa sakta hai; implicit relation $F(x,y)=C$ koi branch choose kiye bina ya information khoe poora solution state karta hai.

Ek solution ki definition hi yahi hai ki uski derivative original ODE reproduce kare; $y^{\prime }=g(x)h(y)$ recover karna exactly yahi confirm karta hai, chahe aapne usse kaise bhi find kiya ho.

$h(y)=y(1-y)$ ke do roots $y=0,1$ hain jo fraction $\frac{A{e}^x}{1+A{e}^x}$ kabhi reach nahi kar sakta; $h(y)=y$ ka single root $y=0$ hai, jo family $A=0$ par recover kar leti hai.

Us point se guzarne wala unique solution constant equilibrium $y\equiv y_0$ hai; aap separate kabhi nahi karte, kyunki $h(y_0)=0$ se divide karna forbidden aur unnecessary hai.

$y^{\prime }=0$, toh $y\equiv$ constant — har horizontal line ek solution hai; separation degenerate ho jaata hai lekin answer immediate hai.

Nahi — uniqueness forbid karta hai ki do solutions ek point share karein, toh non-equilibrium curves $y=1$ ke paas asymptotically aati hain lekin kabhi touch ya cross nahi karti.

Haan, exponential har real $x$ ke liye defined aur smooth hai, toh yeh solution global hai; saare separable solutions aisa nahi hote — implicit wale domain restrict kar sakte hain.

Globally nahi — solve karne par $y=\frac{-1\pm \sqrt{1+4(x^2+C)}}{2}$ milta hai, do branches; initial condition decide karti hai kaunsi branch, aur curve sirf apne extent ke kuch hisse par function ho sakti hai.

Sabhi — yahan $y\equiv 1$ aur $y\equiv 2$ dono alag constant solutions hain, har ek $h$ ka ek root, aur separated family ke alawa har ek report karna zaroori hai.

Haan, yeh $ydy=xdx$ mein separate hota hai; lekin $y=0$ original right side ko blow up kar deta hai, toh koi solution $x$-axis se nahi guzarta aur $h(y)=1/y$ wahan simply undefined hai.