Worked examples — Bernoulli equations — substitution

4.6.5 · D3 · Maths › Ordinary Differential Equations › Bernoulli equations — substitution

Yeh page Bernoulli equations ki exhaustive practice sheet hai. Parent ne machine ko sikhaya; yahan hum use har road pe drive karte hain: positive powers, negative powers, fractions, degenerate cases, ek initial-value problem, ek word problem, aur ek exam trap. Yahan kuch bhi parent se contradict nahi karta — hum bas wider aur slower jaate hain.

Kuch bhi karne se pehle, ek cheez yaad karo. Ek Bernoulli equation ek first-order ODE hai jise is shape mein force kiya ja sakta hai: $\frac{dy}{dx}+P(x)y=Q(x)y^n,n\ne 0,1,$ jahan $P,Q$ sirf $x$ par depend karte hain, aur ek hi cheez hai jo ise linear hone se rokti hai — woh hai power $y^n$. Iski cure (parent mein derive ki gayi) yeh hai:

Yahan "$y^{\prime }$" shorthand hai $\frac{dy}{dx}$ ke liye ("jis rate se $y$ badalta hai jab $x$ move karta hai"), aur $\int \dots dx$ matlab hai "area-accumulating antiderivative". Letter $C$ hamesha woh arbitrary constant hai jo integrate karne par aata hai.

Scenario matrix

Bernoulli problems kuch axes par differ karte hain. Agar tum har row se ek example solve kar sako, toh exam mein aane wali koi bhi Bernoulli equation solve kar sakte ho.

Ab hum har cell hit karte hain.

[!example] Ex 1 — Cell C1 · positive integer power

Solve karo $\frac{dy}{dx}-\frac{y}{x}=x{y}^2.$ ($x>0$ par kaam karo.)

Forecast: Yahan $n=2$ hai, toh $v=y^{1-2}=y^{-1}=1/y$. Kyunki $v$ ek reciprocal hai, guess karo ki final answer "$1/y=(\text{something in }x)$" padhega, na ki directly "$y=\dots$".

1. Identify. $P=-\frac{1}{x},Q=x,n=2.$ Yeh step kyun? Right side par $y^2$ Bernoulli flag karta hai; standard form se match karne par hume recipe ke liye teeno ingredients milte hain.
2. $y^2$ se divide karo: $y^{-2}{y}^{\prime }-\frac{1}{x}{y}^{-1}=x.$ Yeh step kyun? Isse $y^{1-n}=y^{-1}$ expose hota hai aur $y^{-2}{y}^{\prime }$ ka clump banta hai jise chain rule absorb kar lega.
3. Substitute karo $v=y^{-1}$, toh $v^{\prime }=-y^{-2}{y}^{\prime }\Rightarrow y^{-2}{y}^{\prime }=-v^{\prime }.$ Yeh step kyun? Yeh "engine" hai — yeh awkward $y^{-2}{y}^{\prime }$ ko plain $-v^{\prime }$ mein badal deta hai.
4. Linear ODE. Replace karo: $-v^{\prime }-\frac{1}{x}v=x\Rightarrow v^{\prime }+\frac{1}{x}v=-x.$ Yeh step kyun? Yahi $v^{\prime }+(1-n)Pv=(1-n)Q$ mein plug karna hai: $(1-2)(-1/x)=1/x$ aur $(1-2)(x)=-x$. Dono agree karte hain. ✔
5. Integrating factor. $\mu =e^{\int (1/x)dx}=e^{\ln |x|}=x$ ($x>0$ par). Phir $(xv{)}^{\prime }=x\cdot (-x)=-x^2.$ Yeh step kyun? $\mu$ se multiply karna left side ko ek perfect derivative $(\mu v{)}^{\prime }$ bana deta hai, toh hum seedha integrate kar sakte hain. ($x<0$ par hum $\mu =-x$ use karte, jo cancel karne ke baad wahi ODE deta.)
6. Integrate & back-substitute. $xv=-\frac{x^3}{3}+C\Rightarrow v=\frac{C}{x}-\frac{x^2}{3}.$ Kyunki $v=1/y$: $\boxed{\frac{1}{y}=\frac{C}{x}-\frac{x^2}{3}}.$ Yeh step kyun? Hum $(xv{)}^{\prime }=-x^2$ integrate karte hain $xv$ recover karne ke liye, phir substitution $v=1/y$ undo karte hain kyunki recipe ka Step 5 demand karta hai ki hum helper variable se real unknown $y$ par wapas translate karein.

Verify: $C=0$ lo: $1/y=-x^2/3\Rightarrow y=-3/x^2$, toh $y^{\prime }=6/x^3$. $y^{\prime }-y/x$ mein plug karo: $6/x^3-(-3/x^2)/x=6/x^3+3/x^3=9/x^3$. Aur $x{y}^2=x\cdot 9/x^4=9/x^3$. Dono sides match karte hain. ✔

[!example] Ex 2 — Cell C2 · fractional power $0<n<1$

Solve karo $y^{\prime }-y=y^{1/2}$ (parent ka Example 2, hamare fraction case ke roop mein re-verified). Is baar sab recipe steps dikhate hain.

Forecast: $n=\frac{1}{2}$ se $1-n=\frac{1}{2}$ milta hai, toh $v=y^{1/2}=\sqrt{y}$. Naya variable ek square root hai, toh end mein $y$ pane ke liye hum square karenge.

1. Identify. $P=-1,Q=1,n=\frac{1}{2}.$ Yeh step kyun? $y^{\prime }+Py=Q{y}^n$ ke roop mein likha, constants seedha padhte hain.
2. $y^n=y^{1/2}$ se divide karo: $y^{-1/2}{y}^{\prime }-y^{1/2}=1.$ Yeh step kyun? Bilkul Ex 1 ki tarah — $y^{1-n}=y^{1/2}$ expose karo aur $y^{-1/2}{y}^{\prime }$ ka clump banao jise chain rule absorb karega. Hum yeh step skip nahi karte.
3. Substitute karo $v=y^{1/2}$, toh $v^{\prime }=\frac{1}{2}{y}^{-1/2}{y}^{\prime }\Rightarrow y^{-1/2}{y}^{\prime }=2v^{\prime }.$ Yeh step kyun? Yeh awkward $y^{-1/2}{y}^{\prime }$ ko plain $2v^{\prime }$ mein convert karta hai (the "engine").
4. Linear ODE. Step 2 mein replace karo: $2v^{\prime }-v=1\Rightarrow v^{\prime }-\frac{1}{2}v=\frac{1}{2}.$ Yeh step kyun? $2$ se divide karo. Yeh recipe formula $v^{\prime }+(1-n)Pv=(1-n)Q$ se match karta hai jahan $(1-n)=\frac{1}{2}$ dono $P$ aur $Q$ ko multiply karta hai — woh factor jise sab bhool jaate hain. ✔
5. Integrating factor. $\mu =e^{\int -\frac{1}{2}dx}=e^{-x/2}.$ Phir $(e^{-x/2}v{)}^{\prime }=\frac{1}{2}{e}^{-x/2}.$ Yeh step kyun? Left side ko ek single derivative mein badal deta hai.
6. Integrate. $e^{-x/2}v=-e^{-x/2}+C\Rightarrow v=C{e}^{x/2}-1.$ Yeh step kyun? $\int \frac{1}{2}{e}^{-x/2}dx=-e^{-x/2}$; $\mu =e^{-x/2}$ se divide karke $v$ isolate hota hai.
7. Back-substitute $v=\sqrt{y}$: $\boxed{\sqrt{y}=C{e}^{x/2}-1,y=(C{e}^{x/2}-1{)}^2}.$ Yeh step kyun? Helper $v$ ko $y$ mein wapas translate karna zaroori hai; squaring yahan sirf tab valid hai jab $\sqrt{y}\ge 0$ ho — branch note dekho.

Verify: $C=1$ ke saath, $y=(e^{x/2}-1{)}^2$. $x>0$ ke liye $e^{x/2}-1>0$ hai, toh $\sqrt{y}=e^{x/2}-1$ aur $y^{1/2}=e^{x/2}-1$. Phir $y^{\prime }=2(e^{x/2}-1)\cdot \frac{1}{2}{e}^{x/2}=(e^{x/2}-1)e^{x/2}$. Saath hi $y+y^{1/2}=(e^{x/2}-1{)}^2+(e^{x/2}-1)=(e^{x/2}-1)[(e^{x/2}-1)+1]=(e^{x/2}-1)e^{x/2}$. $x>0$ par equal. ✔

[!example] Ex 3 — Cell C3 · genuinely negative power $n<0$

Solve karo $y^{\prime }+y=y^{-2}$ (yahan $n=-2$, sach mein negative index).

Forecast: $n=-2$ se $1-n=1-(-2)=3$ milta hai, toh $v=y^3$ — is baar ek positive power. Kyunki $v=y^3$ hai, final answer "$y^3=\dots$" padhega, aur $y$ pane ke liye hum cube root lenge (odd root, toh koi $\pm$ ambiguity nahi — har real number ka exactly ek real cube root hota hai).

1. Identify. Standard form $y^{\prime }+Py=Q{y}^n$ mein hum padhte hain $P=1,Q=1,n=-2.$ Yeh step kyun? Right side par $y^{-2}$ Bernoulli power hai; $n<0$ allowed hai kyunki $n\ne 0,1$.
2. $y^n=y^{-2}$ se divide karo (yani $y^2$ se multiply karo): $y^2{y}^{\prime }+y^3=1.$ Substitute karo $v=y^3$, toh $v^{\prime }=3y^2{y}^{\prime }\Rightarrow y^2{y}^{\prime }=\frac{1}{3}{v}^{\prime }.$ Yeh step kyun? $y^{1-n}=y^3$ expose karo aur $y^2{y}^{\prime }$ ko $\frac{1}{3}{v}^{\prime }$ mein badlo.
3. Linear ODE. Replace karo: $\frac{1}{3}{v}^{\prime }+v=1\Rightarrow v^{\prime }+3v=3.$ (Recipe $v^{\prime }+(1-n)Pv=(1-n)Q$ ke same, jahan $(1-n)=3$ dono $P$ aur $Q$ ko multiply karta hai.) Yeh step kyun? $3$ se multiply karo; note karo ki $v=y^3$ ek positive power hai — negative-$n$ case ki pehchaan.
4. Integrating factor. $\mu =e^{\int 3dx}=e^{3x}.$ Phir $(e^{3x}v{)}^{\prime }=3e^{3x}.$ Yeh step kyun? $\mu$ se multiply karne par left side ek single derivative $(\mu v{)}^{\prime }$ mein collapse ho jaati hai.
5. Integrate & isolate. $e^{3x}v=e^{3x}+C\Rightarrow v=1+C{e}^{-3x}.$ Yeh step kyun? $\int 3e^{3x}dx=e^{3x}$ integrate karte hain, phir $v$ free karne ke liye $\mu =e^{3x}$ se divide karte hain.
6. Back-substitute $v=y^3$: $\boxed{y^3=1+C{e}^{-3x},y=(1+C{e}^{-3x}{)}^{1/3}.}$ Yeh step kyun? Recipe ka Step 5: helper undo karo. Cube root reals par single-valued hai, toh koi $\pm$ branch nahi aata — Ex 2 ke square-root case se alag.

Verify: $C=0$ ke saath, $y^3=1\Rightarrow y=1$, toh $y^{\prime }=0$. Original: $y^{\prime }+y=0+1=1$ aur $y^{-2}=1^{-2}=1$. Match. ✔ General $C$ ke saath: $y^3=1+C{e}^{-3x}$ differentiate karo taaki $3y^2{y}^{\prime }=-3C{e}^{-3x}$ mile, toh $y^{\prime }=-\frac{C{e}^{-3x}}{y^2}$. Phir $y^{\prime }+y=\frac{-C{e}^{-3x}+y^3}{y^2}=\frac{-C{e}^{-3x}+1+C{e}^{-3x}}{y^2}=\frac{1}{y^2}=y^{-2}$. ✔

[!example] Ex 4 — Cell C4 · initial-value problem

Solve karo $y^{\prime }+\frac{y}{x}=x{y}^2$ with $y(1)=\frac{1}{2}.$ ($x>0$ par kaam karo, jisme $x=1$ hai.)

Forecast: Yeh Ex 1 ka cousin hai lekin plus $y/x$ ke saath. Hume family $1/y=f(x)+Cx$ milegi, aur initial condition ek curve pin karega. $x=1,y=\frac{1}{2}$ par $1/y=2$ hai, toh $C$ ek chhota tidy number aana chahiye.

1. Identify & set up. $P=\frac{1}{x},Q=x,n=2,$ toh $v=y^{-1},v^{\prime }=-y^{-2}{y}^{\prime }.$ Yeh step kyun? Same Bernoulli shape; hum substitution prepare karte hain.
2. Linear ODE. $v^{\prime }+(1-2)\frac{1}{x}v=(1-2)x\Rightarrow v^{\prime }-\frac{1}{x}v=-x.$ Yeh step kyun? Recipe formula mein plug karo.
3. Integrating factor. $\mu =e^{\int -1/xdx}=e^{-\ln |x|}=x^{-1}$ ($x>0$ par). Phir ${\left(\frac{v}{x}\right)}^{\prime }=-1.$ Yeh step kyun? Perfect-derivative trick; $\ln$ mein absolute value hai, $x>0$ par kaam karne se resolve hota hai.
4. General solution. $\frac{v}{x}=-x+C\Rightarrow v=Cx-x^2\Rightarrow \frac{1}{y}=Cx-x^2.$ Yeh step kyun? $(v/x{)}^{\prime }=-1$ integrate karte hain taaki $v/x=-x+C$ mile, $x$ se multiply karke $v$ isolate karte hain, phir $v=1/y$ undo karte hain taaki answer original unknown $y$ mein stated ho.
5. $y(1)=\frac{1}{2}$ apply karo. Toh $1/y=2$: $2=C(1)-1\Rightarrow C=3.$ Yeh step kyun? Initial condition woh ek curve select karti hai jo given point se guzarti hai. $\boxed{\frac{1}{y}=3x-x^2,y=\frac{1}{3x-x^2}}.$

Verify: $x=1$ par: $3(1)-1=2$, toh $y=1/2$. ✔ Differentiate karo: $y^{\prime }=-\frac{3-2x}{(3x-x^2{)}^2}$. $x=1$ par $y^{\prime }+y/x$ check karo: $y^{\prime }=-(1)/(2{)}^2=-1/4$; $y/x=(1/2)/1=1/2$; sum $=1/4$. Aur $x{y}^2=1\cdot (1/2{)}^2=1/4$. Match. ✔

[!example] Ex 5 — Cell C5 · degenerate $n=0$ aur $n=1$

Do mini-problems jo Bernoulli lagte hain par hain nahi. ($x>0$ par kaam karo.)

(a) $n=0$: $y^{\prime }+\frac{2}{x}y=3x.$ Forecast: $n=0$ matlab $y^0=1$, toh "$y^n$" invisible hai — yeh already linear hai. Koi substitution nahi.

1. $\mu =e^{\int 2/xdx}=e^{2\ln |x|}=x^2.$ Kyun? Standard integrating factor; $e^{2\ln |x|}=|x{|}^2=x^2$ dono signs ke liye, toh yahan koi branch issue nahi.
2. $(x^{2}y{)}^{\prime }=3x^3\Rightarrow x^{2}y=\frac{3x^4}{4}+C\Rightarrow y=\frac{3x^2}{4}+\frac{C}{x^2}.$

Verify: $C=0$: $y=\frac{3}{4}{x}^2,y^{\prime }=\frac{3}{2}x$. Phir $y^{\prime }+\frac{2}{x}y=\frac{3}{2}x+\frac{2}{x}\cdot \frac{3}{4}{x}^2=\frac{3}{2}x+\frac{3}{2}x=3x.$ ✔

(b) $n=1$: $y^{\prime }+\frac{1}{x}y=4y.$ Forecast: $n=1$ se $v=y^0=1$ milta (useless!). Iski jagah $y$ collect karo: yeh linear aur separable hai.

1. Rewrite karo: $y^{\prime }=\left(4-\frac{1}{x}\right)y.$ Kyun? Saare $y$-terms ek saath — substitution $1-n=0$ degenerate hai.
2. Separable (dekho Separable Equations): $\frac{dy}{y}=\left(4-\frac{1}{x}\right)dx.$
3. Integrate karo: $\ln |y|=4x-\ln |x|+C_1\Rightarrow \boxed{y=\frac{K{e}^{4x}}{x}}$ ($K=e^{C_1}$ ke saath, aur $\ln |x|$ absolute value phir se).

Verify: $y=K{e}^{4x}/x$, $y^{\prime }=K\frac{4e^{4x}x-e^{4x}}{x^2}=K{e}^{4x}\frac{4x-1}{x^2}$. Phir $y^{\prime }+\frac{1}{x}y=K{e}^{4x}\frac{4x-1}{x^2}+K{e}^{4x}\frac{1}{x^2}=K{e}^{4x}\frac{4x}{x^2}=4\cdot \frac{K{e}^{4x}}{x}=4y.$ ✔

[!example] Ex 6 — Cell C6 · the lost solution $y\equiv 0$

Solve karo $y^{\prime }=y^3$ aur har solution account karo.

Forecast: Yeh Bernoulli hai $P=0,Q=1,n=3$ ke saath. $y^3$ se divide karna silently assume karta hai $y\ne 0$ — lekin $y\equiv 0$ does satisfy karta hai $y^{\prime }=y^3$ (dono sides $0$). Dhyan rakho.

1. Pehle constant solution note karo. $y\equiv 0\Rightarrow y^{\prime }=0=0^3.$ Valid. Ise alag rakhho. Yeh step kyun? Kyunki Step 2 $y^3$ se divide karega aur ise erase kar dega.
2. $y^3$ se divide karo: $y^{-3}{y}^{\prime }=1.$ Substitute karo $v=y^{-2}$ ($1-n=-2$), $v^{\prime }=-2y^{-3}{y}^{\prime }\Rightarrow y^{-3}{y}^{\prime }=-\frac{1}{2}{v}^{\prime }.$ Yeh step kyun? $y^{1-n}$ ka standard exposure.
3. Linear ODE. $-\frac{1}{2}{v}^{\prime }=1\Rightarrow v^{\prime }=-2\Rightarrow v=-2x+C.$ Yeh step kyun? $P=0$ ise trivially integrable banata hai (koi $\ln$ ki zaroorat nahi).
4. Back-substitute $v=y^{-2}$: $\boxed{y^{-2}=C-2x,y=\pm \frac{1}{\sqrt{C-2x}}}\text{plus the singular solution }y\equiv 0.$ Yeh step kyun? Helper $v=y^{-2}$ undo karo. Kyunki $y^{-2}=1/y^2$ $y$ ka sign kho deta hai, $y$ ke liye solve karne par genuinely dono signs $\pm$ aate hain — har ek ek legitimate branch hai, sirf wahan valid jahan $C-2x>0$ ho.

Verify: $C=0$ lo: $y=1/\sqrt{-2x}$ (needs $x<0$), $y=(-2x{)}^{-1/2}$, $y^{\prime }=-\frac{1}{2}(-2x{)}^{-3/2}\cdot (-2)=(-2x{)}^{-3/2}$. Aur $y^3=(-2x{)}^{-3/2}$. Match. ✔ Saath hi $y\equiv 0$ se $0=0$. ✔

[!example] Ex 7 — Cell C7 · real-world word problem (logistic growth)

Ek fish population $y(t)$ (thousands mein) badhti hai, lekin crowding ise limit karti hai. Yeh logistic law follow karti hai: $\frac{dy}{dt}=ry-\frac{r}{K}{y}^2,r=2{\text{yr}}^{-1},K=10\text{thousand}.$ $y(0)=2$ thousand fish se shuru karo. Bernoulli method use karke $y(t)$ dhundho, aur long-run population predict karo.

Forecast: $y^2$ crowding term ise Bernoulli banata hai $n=2$ ke saath. Kyunki fish forever nahi badh sakti, expect karo $y\to K=10$ thousand jab $t\to \infty$. Toh answer $10$ ki taraf flatten hona chahiye.

1. Standard form. $y^{\prime }-2y=-\frac{2}{10}{y}^2=-0.2y^2.$ Toh $P=-2,Q=-0.2,n=2.$ Yeh step kyun? Linear $y$ ko left par isolate karo; $P,Q,n$ identify karo.
2. Substitute karo $v=y^{-1}$. Recipe: $v^{\prime }+(1-2)(-2)v=(1-2)(-0.2)\Rightarrow v^{\prime }+2v=\mathrm{0.2.}$ Yeh step kyun? $(1-n)=-1$ dono $P$ aur $Q$ ko multiply karta hai; signs flip hote hain.
3. Integrating factor. $\mu =e^{\int 2dt}=e^{2t}.$ Phir $(e^{2t}v{)}^{\prime }=0.2e^{2t}.$ Yeh step kyun? Left ko ek derivative mein collapse karo (yahan koi $\ln$ nahi, $P$ constant hai).
4. Integrate. $e^{2t}v=0.1e^{2t}+C\Rightarrow v=0.1+C{e}^{-2t}.$ Yeh step kyun? $\int 0.2e^{2t}dt=0.1e^{2t}$; $v$ isolate karne ke liye $\mu$ se divide karo.
5. Back-substitute $v=1/y$: $\frac{1}{y}=0.1+C{e}^{-2t}.$ $y(0)=2$ apply karo $\Rightarrow \frac{1}{2}=0.1+C\Rightarrow C=\mathrm{0.4.}$ Yeh step kyun? Helper undo karo (populations $1/v$ hain), phir $C$ fix karne ke liye starting count use karo. $\boxed{\frac{1}{y}=0.1+0.4e^{-2t},y(t)=\frac{1}{0.1+0.4e^{-2t}}.}$

Neeche ka figure is solution ko plot karta hai. Horizontal axis: time $t$ years mein. Vertical axis: population $y$ thousands of fish mein. Magenta S-curve orange start point $y(0)=2$ se climb karti hai aur violet dashed line $y=K=10$ (carrying capacity) par flatten ho jaati hai. Qualitatively: growth fast hai jab pond almost empty hai ($ry$ term dominate karta hai), phir throttle hota hai jab crowding — $-\frac{r}{K}{y}^2$ term — bite karta hai. Yeh flattening hi us algebra ka geometric meaning hai jo humne abhi kiya.

Verify (long-run & units): Jab $t\to \infty$, $e^{-2t}\to 0$, toh $\frac{1}{y}\to 0.1\Rightarrow y\to \frac{1}{0.1}=10$ thousand $=K$. ✔ $t=0$ par: $y=1/(0.1+0.4)=1/0.5=2$ thousand, initial count se match karta hai. ✔ Units: $r$ ke units ${\text{yr}}^{-1}$ hain, toh $ry$ ke units (thousand)$\cdot {\text{yr}}^{-1}$ hain, $\frac{dy}{dt}$ se match karte hain; aur $\frac{r}{K}{y}^2$ ke units $\frac{{\text{yr}}^{-1}}{\text{thousand}}\cdot (\text{thousand}{)}^2=(\text{thousand})\cdot {\text{yr}}^{-1}$ bhi hain. Dono terms $y^{\prime }$ ke same units carry karte hain. ✔

[!example] Ex 8 — Cell C8 · exam twist (disguised, plus a Riccati sibling)

(a) Disguised Bernoulli. Solve karo $\frac{dy}{dx}=\frac{y}{x}+\frac{x^2}{y}.$ Yeh standard form bilkul nahi lagta. ($x>0$ par kaam karo.)

Forecast: $\frac{x^2}{y}=x^2{y}^{-1}$ term ka matlab hai $n=-1$. Phir $1-n=2$, toh $v=y^2$ — ek positive power. Expect karo answer "$y^2=\dots$" ke roop mein, phir $y$ recover karne ke liye $\pm$ square root.

1. Standard form mein rearrange karo. $y^{\prime }-\frac{1}{x}y=x^2{y}^{-1}.$ Toh $P=-\frac{1}{x},Q=x^2,n=-1.$ Yeh step kyun? Linear term left par shift karo; $y^{-1}$ ko Bernoulli power ke roop mein pehchano — exam disguise bas un-rearranged layout hai.
2. $y^n=y^{-1}$ se divide karo ($y$ se multiply karo): $y{y}^{\prime }-\frac{1}{x}{y}^2=x^2.$ Substitute karo $v=y^2$, toh $v^{\prime }=2y{y}^{\prime }\Rightarrow y{y}^{\prime }=\frac{1}{2}{v}^{\prime }.$ Yeh step kyun? $y^{1-n}=y^2$ expose karo aur $y{y}^{\prime }$ ko $\frac{1}{2}{v}^{\prime }$ mein badlo.
3. Linear ODE. Replace karo: $\frac{1}{2}{v}^{\prime }-\frac{1}{x}v=x^2\Rightarrow v^{\prime }-\frac{2}{x}v=2x^2.$ (Recipe ke same $(1-n)=2$ ke saath $P$ aur $Q$ ko multiply karte hue.) Yeh step kyun? $\frac{1}{2}$ clear karne ke liye $2$ se multiply karo.
4. Integrating factor. $\mu =e^{\int -2/xdx}=e^{-2\ln |x|}=x^{-2}$ ($x>0$ par). Phir $(x^{-2}v{)}^{\prime }=x^{-2}\cdot 2x^2=2.$ Yeh step kyun? Perfect-derivative form; $\ln$ mein $|x|$ hai, $x>0$ se resolve hota hai.
5. Integrate & back-substitute. $x^{-2}v=2x+C\Rightarrow v=2x^3+C{x}^2.$ $v=y^2$ ke saath: $\boxed{y^2=2x^3+C{x}^2,y=\pm \sqrt{2x^3+C{x}^2}.}$ Yeh step kyun? $(x^{-2}v{)}^{\prime }=2$ integrate karke $x^{-2}v=2x+C$ milta hai, $x^2$ se multiply karke $v$ isolate hota hai, phir $v=y^2$ undo karo. Kyunki $y^2$ $y$ ka sign discard karta hai, $y$ recover karne ke liye dono roots $\pm$ chahiye — har ek ek valid solution branch hai, sirf wahan defined jahan $2x^3+C{x}^2\ge 0$ ho.

Verify: $C=0$: $y^2=2x^3$, lo $y=+\sqrt{2x^3}$ (the $+$ branch), toh $2y{y}^{\prime }=6x^2\Rightarrow y^{\prime }=3x^2/y$. $y/x+x^2/y$ compare karo. $y^2=2x^3$ use karte hue: $y/x=\sqrt{2x^3}/x=\sqrt{2}{x}^{1/2}$ aur $x^2/y=x^2/\sqrt{2x^3}=x^{1/2}/\sqrt{2}$. Sum $=\sqrt{2}{x}^{1/2}+x^{1/2}/\sqrt{2}=\frac{2+1}{\sqrt{2}}{x}^{1/2}=\frac{3}{\sqrt{2}}{x}^{1/2}$. Aur $3x^2/y=3x^2/\sqrt{2x^3}=\frac{3}{\sqrt{2}}{x}^{1/2}$. Match ($-$ branch bhi identically check hota hai $y\to -y$ ke saath). ✔

(b) The Riccati sibling — exams inka pair kyun karte hain. Ek Riccati equation $y^{\prime }=a(x)+b(x)y+c(x)y^2$ mein ek extra constant/known-free term $a(x)$ hota hai, toh yeh Bernoulli nahi hai. Lekin agar tumhe already ek particular solution $y_1(x)$ pata ho, toh substitution $y=y_1+1/z$ ek Riccati ko $z$ mein linear ODE mein badal deta hai — bilkul wahi "rename to linearise" spirit. Jab $a(x)\equiv 0$ ho, Riccati $n=2$ wale Bernoulli mein collapse ho jaata hai.

(Koi naya numeric claim (b) mein nahi — sirf recognition.)

[!recall]- Quick self-test (guess karne ke baad reveal karo)

Connections

• Bernoulli equations — substitution (parent — woh method jise yeh examples exercise karte hain)
• Linear First-Order ODEs — Integrating Factor (har example yahan khatam hota hai)
• Separable Equations (Ex 5b is tarah solve kiya)
• Exact Equations and Integrating Factors
• Substitution Methods in ODEs (wahi rename-to-simplify idea)
• Riccati Equations (Ex 8b — nonlinearity mein ek kadam aage)

Scenario map