Exercises — Separable ODEs — technique, implicit solutions

• (a) Separable. $g(x)=x^3,h(y)=1/y$. RHS ek pure-$x$ piece aur ek pure-$y$ piece ka product hai.
• (b) NOT separable. Ek sum $x-y$ ko (function of $x$)$\times$(function of $y$) ke roop mein nahi likha ja sakta. Ek subtraction ko factor nahi kiya ja sakta.
• (c) Separable. $g(x)=\cos x,h(y)=1+y^2$.
• (d) Separable. Exponent ek sum hai, aur exponent mein sum ek product mein split ho jaata hai: $e^{x-y}=e^x{e}^{-y}$. Toh $g(x)=e^x,h(y)=e^{-y}$.

Yahan $g(x)=1$ aur $h(y)=y(3-y)$ hai. $h(y)=0$ set karo: ya toh $y=0$ ya $3-y=0\Rightarrow y=3$. Toh constant functions $y\equiv 0$ aur $y\equiv 3$ equilibrium solutions hain — jis waqt hum $h(y)$ se divide karenge, yeh mita diye jaayenge, isliye inhe abhi record karte hain.

Pehle lost solutions check karo. Yahan $g(x)=x^3,h(y)=1/y$ hai. Koi $y$ nahi hai jahan $h(y)=0$ ho (kyunki $1/y$ kabhi zero nahi hoti), aur $y=0$ RHS $x^3/y$ ko undefined banata hai — toh $y\equiv 0$ ek solution nahi hai aur solution curves $x$-axis ko kabhi touch nahi karti. Yeh hum isliye note karte hain kyunki $y$ se multiply karna (agli step) warna koi potential constant solution ko silently swallow kar leta. Separate (dono sides ko $y$ se multiply karo, matlab saare $y$ left mein, saare $x$ right mein): $ydy=x^{3}dx.$ Integrate dono sides ko. Har side apna constant laata hai, lekin hum unhe right side par ek single arbitrary constant $C$ mein ikatha kar lete hain: $\frac{y^2}{2}=\frac{x^4}{4}+C.$ Single constant: ek $C$ dono ko absorb karta hai. 2 se multiply karo; kyunki $2C$ abhi bhi bas "koi unknown constant" hai, $2C\to C$ rename karo (ek fresh arbitrary constant): $\boxed{y^2=\frac{x^4}{2}+C.}$ Ise implicit hi rehne dena sabse clean hai; $y$ ke liye solve karne par $\pm \sqrt{}$ aayega (aur sign $y$ ka starting point par sign se determine hoga, kyunki curves $y=0$ cross nahi kar sakti).

Separate: $1+y^2$ se divide karo (kabhi zero nahi, toh koi lost solution nahi): $\frac{dy}{1+y^2}=\cos xdx.$ Integrate. Left side par $\arctan$ kyun? Kyunki $\int \frac{dy}{1+y^2}=\arctan y$ — yahi woh integral hai jiska answer arctan hai. Single arbitrary constant $C$ right side par rakhte hain: $\arctan y=\sin x+C.$ $y$ ke liye solve karo (arctan har jagah invertible hai): $\boxed{y=\tan (\sin x+C).}$ Yeh valid kahaan hai? $\tan$ tab blow up karta hai jab uska argument $\frac{\pi }{2}+k\pi$ ko hit kare. Toh solution sirf tab exist karta hai jab $-\frac{\pi }{2}<\sin x+C<\frac{\pi }{2}(\text{us branch par jo start point contain kare}).$ Initial condition $y(0)=0$ ke liye: $\arctan 0=\sin 0+C\Rightarrow C=0$, jo $y=\tan (\sin x)$ deta hai. Kyunki saare real $x$ ke liye $|\sin x|\le 1<\frac{\pi }{2}\approx 1.571$ hai, argument kabhi $\pm \frac{\pi }{2}$ tak nahi pahunchta, aur yeh particular solution poore $(-\infty ,\infty )$ par defined hai. (Kisi bade $C$ ke liye, jaise $C=1$, argument $\sin x+1$ ka $\frac{\pi }{2}$ se zyada ho sakta hai aur solution phir sirf un $x$-intervals par rehta hai jahan $\sin x+1<\frac{\pi }{2}$ ho.)

Exponent split karo: $e^{x-y}=e^x{e}^{-y}$, toh $g=e^x,h=e^{-y}$. Separate: $e^{-y}$ se divide karo = $e^y$ se multiply karo: $e^{y}dy=e^{x}dx.$ Integrate. $\int e^{y}dy$ exactly $e^y$ kyun deta hai? Kyunki $e^y$ woh ek function hai jo apna khud ka derivative hai: $\frac{d}{dy}{e}^y=e^y$, toh yeh apna khud ka antiderivative bhi hai. Isi tarah $\int e^{x}dx=e^x$. Har integral ek constant laata, lekin — hamesha ki tarah — hum dono ko ek single arbitrary constant $C$ mein merge karte hain aur use sirf right side par rakhte hain: $e^y+({\text{const}}_L)=e^x+({\text{const}}_R)⟹e^y=e^x+C,C={\text{const}}_R-{\text{const}}_L.$ Toh dono integration constants ek dikhaye gaye $C$ mein collapse ho jaate hain. $y$ ke liye solve karo: dono sides ka log lo: $\boxed{y=\ln (e^x+C).}$ Explicit domain. Log sirf positive argument ke liye defined hai, toh chahiye $e^x+C>0$, matlab $e^x>-C$.

• Agar $C\ge 0$: $e^x+C>0$ automatically, toh solution saare $x\in (-\infty ,\infty )$ ke liye defined hai.
• Agar $C<0$: chahiye $e^x>-C\Rightarrow x>\ln (-C)$. Solution sirf $(\ln (-C),\infty )$ par rehta hai, $x\to \ln (-C{)}^{+}$ par $-\infty$ ki taraf blow down karta hua.
• $y(0)=0$ ke liye: $e^0=e^0+C\Rightarrow 1=1+C\Rightarrow C=0$, jo $y=\ln (e^x)=x$ deta hai. Kyunki $C=0\ge 0$, yeh $(-\infty ,\infty )$ poore par valid hai. (Sach mein $y=x$ trivially check karta hai: $y^{\prime }=1=e^{x-x}=e^0=1$.)

Separate & integrate: $\frac{dy}{y}=x^{2}dx\Rightarrow \ln |y|=\frac{x^3}{3}+C$, jahan $C$ integration ka single arbitrary constant hai. Exponentiate: $|y|=e^C{e}^{x^3/3}$, toh $y=A{e}^{x^3/3}$ jahan $A=\pm e^C$ (ek naya nonzero arbitrary constant, $\pm e^C$ ke roop mein defined). $|\cdot |$ hi woh cheez hai jo $A$ ko $\pm$ sign absorb karne deta hai. Apply $y(0)=3$: $3=A{e}^0=A$, toh $A=3$: $\boxed{y=3e^{x^3/3}.}$ Domain: $e^{x^3/3}$ saare real $x$ ke liye finite aur positive hai, toh solution $(-\infty ,\infty )$ par defined hai.

Neeche figure mein horizontal axis $x$ hai aur vertical axis $y$; blue curve solution $y=3e^{x^3/3}$ hai. Yellow dot initial point $(0,3)$ mark karta hai jahan curve $y=3$ se gujarti hai; notice karo ki $x<0$ ke liye curve kitni flat hai ($0$ ki taraf decay karti hai lekin kabhi pahunchti nahi) aur red arrow highlight karta hai ki $x>1$ ke baad explosion kaise hota hai, faster-than-exponential rise (exponent $x^3/3$ khud cube ki tarah badhta hai).

Pehle equilibria (1.2 se): $y\equiv 0$ aur $y\equiv 3$. Separate: $\frac{dy}{y(3-y)}=dx$. Partial fractions. Kyun? Taaki har piece ek log mein integrate ho. Likho: $\frac{1}{y(3-y)}=\frac{A}{y}+\frac{B}{3-y}.$ (Yahan $A,B$ decomposition constants hain, integration constants nahi — inhe algebra fix karta hai, initial data nahi.) Clear karo: $1=A(3-y)+By$. $y=0$ par: $1=3A\Rightarrow A=\frac{1}{3}$. $y=3$ par: $1=3B\Rightarrow B=\frac{1}{3}$. Toh $\frac{1}{y(3-y)}=\frac{1}{3}\left(\frac{1}{y}+\frac{1}{3-y}\right)$. Integrate (right side par ek arbitrary constant $C$): $\frac{1}{3}(\ln |y|-\ln |3-y|)=x+C.$ $\frac{1}{3}$ clear karo har term ko $3$ se multiply karke: $\ln |y|-\ln |3-y|=3x+3C.$ Ab $3C$ abhi bhi bas "koi unknown constant" hai, toh use $C^{\prime }=3C$ rename karo — ek fresh arbitrary constant. Dono logs ko $\ln p-\ln q=\ln \frac{p}{q}$ se combine karo: $\ln \left|\frac{y}{3-y}\right|=3x+C^{\prime }.$ $y$ ke liye solve karo: dono sides exponentiate karo, $\left|\frac{y}{3-y}\right|=e^{C^{\prime }}{e}^{3x}$; $\pm e^{C^{\prime }}\to A$ fold karo (ek aur renamed arbitrary constant) modulus drop ho jaata hai: $\frac{y}{3-y}=A{e}^{3x}.$ Cross-multiply aur $y$ isolate karo: $y=A{e}^{3x}(3-y)\Rightarrow y(1+A{e}^{3x})=3A{e}^{3x}\Rightarrow y=\frac{3A{e}^{3x}}{1+A{e}^{3x}}.$ Apply $y(0)=1$: $1=\frac{3A}{1+A}\Rightarrow 1+A=3A\Rightarrow A=\frac{1}{2}$. $\boxed{y=\frac{\frac{3}{2}{e}^{3x}}{1+\frac{1}{2}{e}^{3x}}=\frac{3e^{3x}}{2+e^{3x}}.}$ Sanity check limit: $x\to \infty$ par, $y\to 3$ (upper equilibrium approach karta hai); $x\to -\infty$ par, $y\to 0$.

Neeche figure mein horizontal axis $x$ hai aur vertical axis $y$; green curve hamara solution $y=\frac{3e^{3x}}{2+e^{3x}}$ hai. Yellow dashed line $y=3$ par aur red dashed line $y=0$ par dono equilibrium solutions hain: green curve blue dot $(0,1)$ se start hoti hai, un dono flat lines ke beech squeezed hai, S-shape mein rise karti hai, aur $y=3$ ke just neeche level off ho jaati hai — visually dikhata hai ki $y=0$ aur $y=3$ ek "floor" aur "ceiling" ki tarah act karte hain jinhe solution approach kar sakta hai lekin kabhi cross nahi kar sakta.

Equilibrium: $h(y)=y-2=0\Rightarrow y\equiv 2$ ek solution hai (iska slope $0=(0)\cos x$ hai). Use record karo. Generic curves: $y-2$ se divide karo: $\frac{dy}{y-2}=\cos xdx\Rightarrow \ln |y-2|=\sin x+C,$ jahan $C$ integration ka single arbitrary constant hai. Exponentiate karo: $y-2=A{e}^{\sin x}$ (jahan $A=\pm e^C$, ek naya nonzero arbitrary constant): $\boxed{y=2+A{e}^{\sin x}(A\ne 0),\text{plus the equilibrium }y\equiv 2.}$ Note karo $A=0$ wapis $y\equiv 2$ deta hai, toh hum kisi bhi real $A$ ki allow kar sakte hain aur ek hi formula sab cover karta hai — lekin sirf isliye kyunki humne consciously check kiya.

Equilibrium: $y^2=0\Rightarrow y\equiv 0$ ek solution hai ($y(0)=1$ fit nahi hoga, lekin record karo). Separate: $\frac{dy}{y^2}=dx\Rightarrow -\frac{1}{y}=x+C$, jahan $C$ integration ka arbitrary constant hai. Apply $y(0)=1$: $-1=0+C\Rightarrow C=-1$. Toh $-\frac{1}{y}=x-1\Rightarrow y=\frac{1}{1-x}$. $\boxed{y=\frac{1}{1-x}.}$ Interval of validity: denominator $1-x=0$ at $x=1$, jahan $y\to \infty$ (finite-time blow-up!). $(0,1)$ se gujarne wala solution sirf $(-\infty ,1)$ par rehta hai — woh sabse bada interval jo $x=0$ contain kare aur singularity se bache. Yahi classic lesson hai: ek bilkul smooth dikhne wala ODE ek aisa solution produce kar sakta hai jo finite $x$ mein infinity tak escape kar jaata hai.

Separable form mein laao. RHS factor karo: $y+xy=y(1+x)$. $x$ se divide karo: $\frac{dy}{dx}=\frac{y(1+x)}{x}=y\cdot \frac{1+x}{x}.$ Ab $g(x)=\frac{1+x}{x}=\frac{1}{x}+1$ aur $h(y)=y$ — separable! Equilibrium: $y\equiv 0$. Separate & integrate (single arbitrary constant $C$): $\frac{dy}{y}=\left(\frac{1}{x}+1\right)dx\Rightarrow \ln |y|=\ln |x|+x+C.$ $y$ ke liye solve karo: exponentiate karo, $|y|=e^C|x|e^x$, aur kyunki $x>0$ (toh $|x|=x$), $\pm e^C\to A$ fold karo: $\boxed{y=Ax{e}^x(A\text{ any constant, }A=0\text{ gives }y\equiv 0).}$

Us singularity ka dhyan rakho jisse tum divide karne wale ho. RHS undefined hai jab $2y-1=0$, matlab $y=\frac{1}{2}$. Wahaan slope $y^{\prime }$ infinite hoga (ek vertical tangent), toh ek genuine solution curve $y=\frac{1}{2}$ ko sirf ek isolated point par touch kar sakta hai jahan vertical tangent ho, aur $x$ ke function ke roop mein kabhi cross nahi kar sakta. Domain ke liye yeh yaad rakho. Separate: $(2y-1)dy=(3x^2+1)dx$. Integrate (single arbitrary constant $C$): $y^2-y=x^3+x+C$. $\boxed{y^2-y-x^3-x=C.}$ Kyun rokein? $y$ isolate karne ke liye quadratic formula chahiye jisme sign ambiguity hai — implicit form honest general solution hai. Square complete karne par, ${\left(y-\frac{1}{2}\right)}^2=x^3+x+C+\frac{1}{4}$, toh $y=\frac{1}{2}\pm \sqrt{x^3+x+C+\frac{1}{4}}.$ Kaun sa sign / kahaan valid? Initial condition sign pick karti hai: agar $y_0>\frac{1}{2}$ toh "$+$" lo (poora branch $\frac{1}{2}$ ke upar rehta hai); agar $y_0<\frac{1}{2}$ toh "$-$" lo (neeche rehta hai). Solution sirf wahaan exist karta hai jahan radicand $x^3+x+C+\frac{1}{4}\ge 0$ ho; jis point par yeh $0$ hit kare wahan curve $y=\frac{1}{2}$ tak pahunche vertical tangent ke saath aur function-solution stop ho jaaye. Toh interval of validity woh $x$-interval hai (jisme $x_0$ ho) jis par $x^3+x+C+\frac{1}{4}>0$ ho. Verify implicit differentiation se: $2y{y}^{\prime }-y^{\prime }-3x^2-1=0\Rightarrow y^{\prime }(2y-1)=3x^2+1\Rightarrow y^{\prime }=\frac{3x^2+1}{2y-1}.$ ✓

Usual sense mein koi equilibrium nahi, lekin note karo ki $y=0$ RHS $x/y$ ko undefined banata hai — solutions $x$-axis cross nahi karte. Separate: $ydy=xdx$. Integrate (single arbitrary constant): $\frac{y^2}{2}=\frac{x^2}{2}+K$. $2$ se multiply karo aur $2K\to C$ rename karo (ek fresh arbitrary constant): $\boxed{y^2-x^2=C.}$ Yeh hyperbolas hain (ya unka degenerate limit). Hume $C$ ke teeno signs handle karne hain, aur har mein $y_0$ ka sign bhi:

Case $C>0$. Tab $y^2=x^2+C\ge C>0$, toh $y$ kabhi zero nahi. Solve karne par, $y=\pm \sqrt{x^2+C}$ — do alag branches:

• upper branch $y=+\sqrt{x^2+C}$ poori tarah $y>0$ mein rehti hai;
• lower branch $y=-\sqrt{x^2+C}$ poori tarah $y<0$ mein rehti hai. Ek solution us branch par rehta hai jo $y_0$ ke sign se match kare ($y=0$ cross nahi ho sakta). Dono branches saare $x\in (-\infty ,\infty )$ ke liye defined hain (radicand hamesha positive hai).

Case $C<0$. Ab $y^2=x^2+C$ ko chahiye $x^2\ge -C$, matlab $|x|\ge \sqrt{-C}$. Hyperbola left/right open hai: do pieces, ek $x\le -\sqrt{-C}$ par aur ek $x\ge +\sqrt{-C}$ par, beech mein ek gap jahan koi real $y$ exist nahi karta. Phir bhi $y=\pm \sqrt{x^2+C}$, aur $y_0$ ka sign upper ($+$) ya lower ($-$) select karta hai. Har solution sirf apne $x$-interval par (do arms mein se ek) defined hai, aur endpoint $|x|=\sqrt{-C}$ par curve $y=0$ tak vertical tangent ke saath pahunchti hai, toh $x$ ke function ke roop mein wahan rok jaati hai.

Case $C=0$ (degenerate). Tab $y^2=x^2\Rightarrow y=\pm x$ — do straight asymptote lines. Kyunki solutions origin se nahi guzar sakte (RHS $y=0$ par undefined), har ek sirf $x>0$ ya $x<0$ par valid hai, jo char half-line solutions deta hai: $y=x(x>0)$, $y=x(x<0)$, $y=-x(x>0)$, $y=-x(x<0)$. $y_0$ ka sign aur $x_0$ ka sign milke decide karte hain ki tum kis half-line par ho.

"$y_0$ ke saare signs" ka summary: $y_0$ ka sign har case mein upper (+) ya lower (−) branch decide karta hai; value $C=y_0^2-x_0^2$ shape decide karti hai (agar $C>0$ toh up/down open, $C<0$ toh left/right open, $C=0$ toh degenerate lines); aur koi bhi solution kabhi $y=0$ cross nahi karta.

Teeno cases neeche ek saath draw hain.

Equilibrium: $y\equiv 0$ ($y(1)=1$ meet nahi karega, lekin note karo). Separate: $\frac{dy}{y^2}=\frac{dx}{x(x+1)}$. Right side par partial fractions. Denominator $x(x+1)$ ke do distinct linear factors hain, toh har factor ke liye ek term mein split karo: $\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}(\text{check: }\frac{1}{x}-\frac{1}{x+1}=\frac{(x+1)-x}{x(x+1)}=\frac{1}{x(x+1)}✓).$ Integrate (single arbitrary constant $C$): $-\frac{1}{y}=\ln |x|-\ln |x+1|+C=\ln \left|\frac{x}{x+1}\right|+C.$ Apply $y(1)=1$: LHS $=-1$; RHS $=\ln \frac{1}{2}+C$. Toh $C=-1-\ln \frac{1}{2}=-1+\ln 2$. Toh $-\frac{1}{y}=\ln \left|\frac{x}{x+1}\right|-1+\ln 2=\ln \left|\frac{2x}{x+1}\right|-1$. $y$ ke liye solve karo: $\boxed{y=\frac{1}{1-\ln \left|\frac{2x}{x+1}\right|}.}$ Start point $x=1$ ke aas paas explicit interval of validity. Teen cheezein hold karni chahiye:

1. ODE ki khud ki singularities $x=0$ aur $x=-1$ se door raho (jahan $x(x+1)=0$ ho). Kyunki $1>0$, hame chahiye $x>0$, jo $0$ par ya uske baayein sab kuch rule out karta hai.
2. Log ka argument positive rakho: $x>0$ ke liye $\frac{2x}{x+1}>0$ automatically, toh yahan koi extra constraint nahi.
3. $y$ ke denominator ko nonzero rakho: $1-\ln \frac{2x}{x+1}=0\Rightarrow \ln \frac{2x}{x+1}=1\Rightarrow \frac{2x}{x+1}=e$. Solve karne par: $2x=e(x+1)\Rightarrow x(2-e)=e\Rightarrow x=\frac{e}{2-e}\approx \frac{2.718}{-0.718}\approx -3.79$. Yeh root negative hai, isliye hamari region $x>0$ mein nahi, toh $x>0$ ke liye denominator kabhi vanish nahi karta.

Combine karne par, woh sabse bada interval jo $x=1$ contain kare aur jis par saari conditions hold karein: $\boxed{(0,\infty ).}$ Important baat yeh method hai: two-distinct-factor partial fractions ne $x(x+1)$ ko untangle kiya, phir singularity-plus-denominator analysis ne explicit interval $(0,\infty )$ pin kiya.

$x+y$, $e^{x+y}$, $x-y$, $\cos x(1+y^2)$ mein se kaun separable hain? $e^{x+y}=e^x{e}^y$ ✓ aur $\cos x(1+y^2)$ ✓ separable hain; bare sums $x+y$ aur $x-y$ nahi hain (exponent ke bahar $x$ aur $y$ ke beech $\pm$ hona factoring ko block karta hai).

(1) Kya $h(y)$ se divide karne par koi equilibrium $y\equiv y_0$ jahan $h(y_0)=0$ delete to nahi ho gayi? (2) Interval of validity kya hai — denominator kahaan vanish karta hai, log non-positive kab hota hai, ya solution blow up kab karta hai?

$C$ (ya $C^{\prime }$, $A$) kya represent karta hai? Ek arbitrary constant of integration — ek abhi-tak-unknown fixed number jo isliye appear hota hai kyunki antiderivatives sirf ek constant tak determined hote hain. Ise reshape karna (multiply karna, exponentiate karna) sirf use rename karta hai; ek first-order ODE ke liye hamesha exactly ek aisa constant hota hai, jo ek initial condition se pin hota hai.