Exercises — Bernoulli equations — substitution

Ek Bernoulli equation ka form hona chahiye $y^{\prime }+P(x)y=Q(x)y^n$ jahan nonlinearity sirf ek power $y^n$ mein ho, aur $n\ne 0,1$.

(a) Bernoulli: $P=3$, $Q=5$, $n=4$. ✔ (b) Bernoulli: $P=\frac{2}{x}$, $Q=x^3$, $n=-2$. ✔ (negative $n$ allowed hai — $n\in \mathbb{R}$) (c) Bernoulli nahi interesting sense mein: yahan $n=0$ hai (right side $x=x\cdot y^0$), jo ki already linear hai. Koi substitution nahi chahiye. (d) Bernoulli: $P=\sin x$, $Q=1$, $n=\frac{1}{3}$. ✔

(a) $1-n=1-4=-3$, toh $v=y^{-3}$. (b) $1-n=1-(-2)=3$, toh $v=y^3$. (d) $1-n=1-\frac{1}{3}=\frac{2}{3}$, toh $v=y^{2/3}$.

Identify karo: $P=\frac{1}{x}$, $Q=x$, $n=3$, toh $1-n=-2$ aur $v=y^{-2}$.

Master result: $v^{\prime }+(1-n)Pv=(1-n)Q$ deta hai $v^{\prime }+(-2)\frac{1}{x}v=(-2)x⟹v^{\prime }-\frac{2}{x}v=-2x.$ Integrating factor: $\mu =e^{\int -2/xdx}=e^{-2\ln x}=x^{-2}.$ Multiply karo: $(x^{-2}v{)}^{\prime }=-2x\cdot x^{-2}=-\frac{2}{x}.$ Integrate karo: $x^{-2}v=-2\ln |x|+C\Rightarrow v=x^2(C-2\ln |x|).$ Back-substitute karo $v=y^{-2}$: $\boxed{\frac{1}{y^2}=x^2(C-2\ln |x|)}$

Identify karo: $P=-1$, $Q=x$, $n=2$, $1-n=-1$, $v=y^{-1}$. $v^{\prime }+(-1)(-1)v=(-1)x⟹v^{\prime }+v=-x.$ IF: $\mu =e^{\int 1dx}=e^x.$ Tab $(e^{x}v{)}^{\prime }=-x{e}^x.$ Integration by parts: $\int -x{e}^{x}dx=-(x-1)e^x+C=(1-x)e^x+C.$ Toh $e^{x}v=(1-x)e^x+C\Rightarrow v=(1-x)+C{e}^{-x}.$ Back-substitute karo $v=1/y$: $\boxed{\frac{1}{y}=1-x+C{e}^{-x}}$

Identify karo: $P=1$, $Q=\frac{1}{2}$, $n=3$, $1-n=-2$, $v=y^{-2}$. $v^{\prime }+(-2)(1)v=(-2)\frac{1}{2}⟹v^{\prime }-2v=-1.$ IF: $\mu =e^{-2x}$. Tab $(e^{-2x}v{)}^{\prime }=-e^{-2x}.$ Integrate karo: $e^{-2x}v=\frac{1}{2}{e}^{-2x}+C\Rightarrow v=\frac{1}{2}+C{e}^{2x}.$ IC apply karo: $y(0)=1\Rightarrow v(0)=1^{-2}=1$. Toh $1=\frac{1}{2}+C\Rightarrow C=\frac{1}{2}.$ Is tarah $v=\frac{1}{2}+\frac{1}{2}{e}^{2x}=\frac{1}{2}(1+e^{2x}).$ Back-substitute karo $v=y^{-2}$: $\boxed{y^2=\frac{2}{1+e^{2x}}}⟹y=\sqrt{\frac{2}{1+e^{2x}}}$ (positive root kyunki $y(0)=1>0$). Sanity check: $x=0$ pe, $y^2=\frac{2}{2}=1$. ✔

Standard form mein rearrange karo: $y^{\prime }-(\tan x)y=(\cos x)y^4.$ Toh $P=-\tan x$, $Q=\cos x$, $n=4$, $1-n=-3$, $v=y^{-3}$. $v^{\prime }+(-3)(-\tan x)v=(-3)\cos x⟹v^{\prime }+3\tan xv=-3\cos x.$ IF: $\mu =e^{\int 3\tan xdx}=e^{-3\ln |\cos x|}=|\cos x{|}^{-3}={\sec }^{3}x$ (ek interval pe jahan $\cos x>0$). Tab $({\sec }^{3}xv{)}^{\prime }=-3\cos x{\sec }^{3}x=-3{\sec }^{2}x.$ Integrate karo: ${\sec }^{3}xv=-3\tan x+C\Rightarrow v={\cos }^{3}x(C-3\tan x).$ Back-substitute karo $v=y^{-3}$: $\boxed{y^{-3}={\cos }^{3}x(C-3\tan x)}$ Lost solution: $y^4$ se divide karne pe assume kiya $y\ne 0$. Kyunki $n=4>0$ hai, $y\equiv 0$ original equation satisfy karta hai ($0=0$) aur yeh ek genuine singular solution hai jo kisi finite $C$ se capture nahi hoti.

$y=y_1+u$ substitute kyun karein? Ek Riccati equation ke liye, jab ek solution $y_1$ pata ho, toh $y=y_1+u$ likhne se constant clutter cancel ho jaata hai aur ek Bernoulli equation $n=2$ ke saath $u$ mein milta hai (dekho Riccati Equations).

$y=1+u$ ke saath, $y^{\prime }=u^{\prime }$. Substitute karo: $u^{\prime }=1+x-(1+2x)(1+u)+x(1+u{)}^2.$ Right side expand karo: $1+x-(1+2x)-(1+2x)u+x(1+2u+u^2).$ Constant terms: $1+x-1-2x+x=0.$ Achha — $y_1=1$ solution tha, toh woh cancel ho jaate hain. Linear-in-$u$ terms: $-(1+2x)u+2xu=-u.$ Quadratic: $x{u}^2.$ Toh $u^{\prime }=-u+x{u}^2$, yaani $u^{\prime }+u=x{u}^2.$ Yeh Bernoulli hai $P=1$, $Q=x$, $n=2$ ke saath — bilkul Exercise 2.2 jaisa! Wahan se $\frac{1}{u}=1-x+C{e}^{-x}.$ $y$ pe wapas jaao: $u=y-1$, toh $\boxed{\frac{1}{y-1}=1-x+C{e}^{-x}}$

Roles flip kyun karein? Jaise likha hai, $y$-powers denominator mein hain — standard shape nahi hai. Invert karo: $\frac{dx}{dy}=\frac{2x-x^{2}y}{y}.$ $\frac{dx}{dy}=\frac{2}{y}x-y{x}^2⟹\frac{dx}{dy}-\frac{2}{y}x=-y{x}^2.$ Ab yeh $x(y)$ mein Bernoulli hai: independent variable $y$, dependent $x$, $P(y)=-\frac{2}{y}$, $Q(y)=-y$, $n=2$, $1-n=-1$, $v=x^{-1}$ ke saath. $v^{\prime }+(-1)\left(-\frac{2}{y}\right)v=(-1)(-y)⟹\frac{dv}{dy}+\frac{2}{y}v=y.$ IF: $\mu =e^{\int 2/ydy}=e^{2\ln |y|}=y^2.$ Tab $(y^{2}v{)}^{\prime }=y^3.$ Integrate karo: $y^{2}v=\frac{y^4}{4}+C\Rightarrow v=\frac{y^2}{4}+\frac{C}{y^2}.$ Back-substitute karo $v=1/x$: $\boxed{\frac{1}{x}=\frac{y^2}{4}+\frac{C}{y^2}}$

$P=\frac{a}{x}$, $Q=b{x}^k$. $v=y^{1-n}$ ke saath: $v^{\prime }+(1-n)\frac{a}{x}v=(1-n)b{x}^k.$ Maano $m=(1-n)a$ aur $c=(1-n)b$. IF: $\mu =e^{\int m/xdx}=x^m.$ Tab $(x^{m}v{)}^{\prime }=c{x}^{k+m}.$ Generic case ($k+m\ne -1$): $x^{m}v=\frac{c{x}^{k+m+1}}{k+m+1}+C⟹v=\frac{c{x}^{k+1}}{k+m+1}+C{x}^{-m}.$ Back-substitute karo: $\boxed{y^{1-n}=\frac{(1-n)b{x}^{k+1}}{k+(1-n)a+1}+C{x}^{-(1-n)a}}$ Logarithmic case: agar $k+m=-1$ ho, yaani $k+(1-n)a=-1$, toh integral $\int c{x}^{-1}dx=c\ln |x|$ ho jaata hai, aur milta hai $y^{1-n}=x^{-m}(c\ln |x|+C).$

Standard form: $y^{\prime }-y=-t{y}^2$, Bernoulli hai $P=-1$, $Q=-t$, $n=2$, $1-n=-1$, $v=y^{-1}$ ke saath. $v^{\prime }+(-1)(-1)v=(-1)(-t)⟹v^{\prime }+v=t.$ IF: $\mu =e^t$. Tab $(e^{t}v{)}^{\prime }=t{e}^t.$ Integration by parts: $\int t{e}^{t}dt=(t-1)e^t+C.$ Toh $e^{t}v=(t-1)e^t+C\Rightarrow v=(t-1)+C{e}^{-t}.$ IC: $y(0)=2\Rightarrow v(0)=\frac{1}{2}$. Toh $\frac{1}{2}=(0-1)+C=-1+C\Rightarrow C=\frac{3}{2}.$ Is tarah $v=t-1+\frac{3}{2}{e}^{-t}.$ Back-substitute karo $v=1/y$: $\boxed{y(t)=\frac{1}{t-1+\frac{3}{2}{e}^{-t}}}$ Long-time behaviour: jab $t\to \infty$, $e^{-t}\to 0$ toh $v\approx t-1\to \infty$, isliye $y\approx \frac{1}{t}\to 0$. Tightening carrying capacity ($-t{y}^2$ term barhta hai) population ko $1/t$ ki tarah extinction ki taraf drive karta hai. Neeche curve dekho.

Jab $n\to 1$, $1-n\to 0$, toh $v=y^{1-n}\to y^0=1$ — ek constant, jo koi information carry nahi karta. Linear ODE $v^{\prime }+(1-n)Pv=(1-n)Q$ limit mein $0=0$ ban jaata hai: dono sides vanish ho jaate hain. Toh machinery collapse ho jaati hai.

$n=1$ pe actually kya hota hai: original equation $y^{\prime }+Py=Q{y}^1$ hai $y^{\prime }+(P-Q)y=0,$ jo ki separable / linear-homogeneous hai: $\frac{dy}{y}=(Q-P)dx$, deta hai $y=C{e}^{\int (Q-P)dx}$. Koi substitution nahi chahiye — directly solve karo. Yahi reason hai kyun definition $n=1$ ko exclude karti hai (dekho Separable Equations).