Exercises — Cauchy-Euler (Equidimensional) equation

Pehchaan ka tarika yeh hai: $x$ ki power = uske saath aane wali derivative ka order. Har term dekho.

• (a) $x^2{y}^{\prime \prime }$ (power 2, order 2 ✓), $5x{y}^{\prime }$ (power 1, order 1 ✓), $3y$ (power 0, order 0 ✓). Haan. Yahan $a=1,b=5,c=3$, toh indicial: $m(m-1)+5m+3=0⟹m^2+4m+3=0.$
• (b) Middle term $5y^{\prime }$ hai — $x$ ki power $0$ hai lekin derivative order $1$ hai. Mismatch. Cauchy-Euler nahi.
• (c) $3x^2{y}^{\prime \prime }$ (✓), $-x{y}^{\prime }$ (✓), $y$ (✓). Haan. $a=3,b=-1,c=1$: $3m(m-1)-m+1=0⟹3m^2-4m+1=0.$

Yahan $a=1,b=-7,c=16$. Indicial: $m(m-1)-7m+16=0⟹m^2-8m+16=0.$ Discriminant hai $b^{\prime 2}-4a^{\prime }{c}^{\prime }=(-8{)}^2-4(1)(16)=64-64=0$. Zero discriminant ka matlab hai ek repeated root — Case 2, toh ek $\ln x$ aane wala hai.

$a=1,b=-2,c=-4$. Indicial: $m(m-1)-2m-4=m^2-3m-4=0$. Factor karo: $(m-4)(m+1)=0\Rightarrow m=4,-1$. Distinct real roots → do independent powers: $\boxed{y=C_1{x}^4+C_2{x}^{-1}}$

$a=1,b=-3,c=4$. Indicial: $m(m-1)-3m+4=m^2-4m+4=(m-2{)}^2=0\Rightarrow m=2$ (double). Ek power $x^2$; doosra solution ek $\ln x$ leta hai ($t=\ln x$ wali duniya mein $t{e}^{mt}$ solution se): $\boxed{y=(C_1+C_2\ln x)x^2}$

$a=1,b=1,c=1$. Indicial: $m(m-1)+m+1=m^2+1=0\Rightarrow m=\pm i$. Toh $\alpha =0,\beta =1$. Kyunki $x^{\pm i}=e^{\pm i\ln x}$ hai, Euler's Formula real trig mein convert karta hai: $\boxed{y=C_1\cos (\ln x)+C_2\sin (\ln x)}$

$a=2,b=3,c=-1$. Indicial: $2m(m-1)+3m-1=2m^2+m-1=0$. Quadratic formula: $m=\frac{-1\pm \sqrt{1+8}}{4}=\frac{-1\pm 3}{4}$, jisse $m=\frac{1}{2}$ aur $m=-1$ milte hain. $\boxed{y=C_1{x}^{1/2}+C_2{x}^{-1}}$

$a=1,b=5,c=4$. Indicial: $m(m-1)+5m+4=m^2+4m+4=(m+2{)}^2=0\Rightarrow m=-2$ (repeated). General solution: $y=(C_1+C_2\ln x)x^{-2}$. Ab conditions lagao. $x=1$ par, $\ln 1=0$, toh $y(1)=C_1=3$. Product rule se differentiate karo: $y^{\prime }=C_2\cdot \frac{1}{x}\cdot x^{-2}+(C_1+C_2\ln x)\cdot (-2)x^{-3}=[C_2-2(C_1+C_2\ln x)]x^{-3}.$ $x=1$ par: $y^{\prime }(1)=C_2-2C_1=C_2-6=-11\Rightarrow C_2=-5$. $\boxed{y=(3-5\ln x)x^{-2}}$

$a=1,b=1,c=4$. Indicial: $m(m-1)+m+4=m^2+4=0\Rightarrow m=\pm 2i$. Toh $\alpha =0,\beta =2$. $\boxed{y=C_1\cos (2\ln x)+C_2\sin (2\ln x)}$ Behaviour. Kyunki $\alpha =0$ hai, aage koi growing/decaying power factor $x^{\alpha }$ nahi hai — amplitude constant rehti hai. Argument hai $2\ln x$. Jab $x\to \infty$, $\ln x\to \infty$ slowly hota hai, toh curve oscillate karta rehta hai lekin oscillations stretch out hoti jaati hain (har cycle ek wider range of $x$ span karti hai). Jab $x\to 0^{+}$, $\ln x\to -\infty$ hota hai, toh curve infinitely many times oscillate karta hai, waves ek doosre ke paas simeetti jaati hain — kabhi koi limit nahi aati. Chalkboard sketch dekho.

$a=1,b=-5,c=8$. Indicial: $m(m-1)-5m+8=m^2-6m+8=(m-2)(m-4)=0\Rightarrow m=2,4$. $x>0$ par hum $y=C_1{x}^2+C_2{x}^4$ likhte. Lekin yahan $x<0$ hai. $x^{1/2}$ ya $\ln x$ jaisi power negative $x$ ke liye undefined hogi, toh hum $|x|$ se apna bachaav karte hain: solutions $|x{|}^m$ se bante hain (aur zarurat padne par $\ln |x|$ se). Yahan exponents $2,4$ even integers hain, toh $x^2=|x{|}^2$ aur $x^4=|x{|}^4$ waise bhi — lekin safe general form hai: $\boxed{y=C_1|x{|}^2+C_2|x{|}^4=C_1{x}^2+C_2{x}^4}$ $|x{|}^m$ likhne ki aadat us waqt matter karti hai jab exponent fractional ho ya $\ln$ aaye.

Substitute $t=\ln x$. Parent note se, $x{y}^{\prime }=\dot{y}$ aur $x^2{y}^{\prime \prime }=\ddot{y}-\dot{y}$ (dots matlab $d/dt$). Yahan $a=1,b=-1,c=1$, toh transformed constant-coefficient ODE hai $\ddot{y}+(b-a)\dot{y}+cy=\ddot{y}-2\dot{y}+y=t.$ (Dhyan do right side $\ln x$ bana $t$.) Homogeneous part: characteristic $m^2-2m+1=(m-1{)}^2=0\Rightarrow m=1$ (repeated), toh $y_h=(C_1+C_{2}t)e^t=(C_1+C_2\ln x)x$. Particular part undetermined coefficients se: $y_p=At+B$ try karo (pehli degree ka polynomial right side $t$ se match karta hai; $e^t$ se clash nahi karta). Phir ${\dot{y}}_p=A,{\ddot{y}}_p=0$: $0-2A+(At+B)=t\Rightarrow At+(B-2A)=t.$ Match karo: $A=1$ aur $B-2A=0\Rightarrow B=2$. Toh $y_p=t+2=\ln x+2$. $\boxed{y=(C_1+C_2\ln x)x+\ln x+2}$

Homogeneous: $a=1,b=-1,c=-3$. Indicial: $m(m-1)-m-3=m^2-2m-3=(m-3)(m+1)=0\Rightarrow m=3,-1$. $y_h=C_1{x}^3+C_2{x}^{-1}.$ Particular: right side $x^2$ hai. Kyunki $2$ indicial equation ka root nahi hai ($m=3,-1$), power $x^2$ homogeneous solution nahi hai, toh plain guess $y_p=A{x}^2$ safe hai (koi resonance nahi). Phir $y_p^{\prime }=2Ax,y_p^{\prime \prime }=2A$. Substitute karo: $x^2(2A)-x(2Ax)-3(A{x}^2)=(2A-2A-3A)x^2=-3A{x}^2=x^2.$ Toh $-3A=1\Rightarrow A=-\frac{1}{3}$, jisse $y_p=-\frac{1}{3}{x}^2$ milta hai. $\boxed{y=C_1{x}^3+C_2{x}^{-1}-\frac{1}{3}{x}^2}$

Homogeneous: $a=1,b=-1,c=1$. Indicial: $m(m-1)-m+1=m^2-2m+1=(m-1{)}^2=0\Rightarrow m=1$ (repeated). $y_h=(C_1+C_2\ln x)x.$ Particular: right side $x^1$ hai, aur $m=1$ indicial equation ka double root hai — double resonance. $t$-world mein yeh waisa hai jaise $e^t$ ko $(m-1{)}^2$ ke against force karo, jisme $t^2{e}^t$ chahiye. Wapas translate karte hue, hum try karte hain $y_p=Ax(\ln x{)}^2.$ $t=\ln x$ mein kaam karna easy hai: transformed equation hai $\ddot{y}-2\dot{y}+y=e^t$ (kyunki $x=e^t$). Try karo $y_p=A{t}^2{e}^t$. $(D-1{)}^2[t^2{e}^t]=2A{e}^t$ use karke (do derivatives polynomial ko uske constant $2A$ tak le jaati hain), hamein $2A=1\Rightarrow A=\frac{1}{2}$ chahiye. Toh $y_p=\frac{1}{2}{t}^2{e}^t=\frac{1}{2}(\ln x{)}^{2}x$. $\boxed{y=(C_1+C_2\ln x)x+\frac{1}{2}(\ln x{)}^{2}x}$

Third order ke liye, $y=x^m$ mein $x^3{y}^{\prime \prime \prime }=m(m-1)(m-2)x^m$, $x^2{y}^{\prime \prime }=m(m-1)x^m$, $x{y}^{\prime }=m{x}^m$, $y=x^m$ chahiye. Indicial: $m(m-1)(m-2)+2m(m-1)-m+1=0.$ Expand karo: $m(m-1)(m-2)=m^3-3m^2+2m$; $2m(m-1)=2m^2-2m$. $-m+1$ ke saath sum: $m^3-3m^2+2m+2m^2-2m-m+1=m^3-m^2-m+1=0.$ Grouping se factor karo: $m^2(m-1)-(m-1)=(m-1)(m^2-1)=(m-1{)}^2(m+1)=0$. Roots: $m=1$ (double), $m=-1$ (simple). Double root $\to$ pair $x^1$ aur $(\ln x)x^1$; simple root $\to x^{-1}$: $\boxed{y=(C_1+C_2\ln x)x+C_3{x}^{-1}}$

Solution form $x^{\alpha }[\dots \cos (\beta \ln x)\dots ]$ humein batata hai ki roots hain $m=\alpha \pm i\beta$ jahan $\alpha =3,\beta =1$ hai, yani $m=3\pm i$. Un roots ke saath indicial polynomial hai $(m-(3+i))(m-(3-i))=(m-3{)}^2+1=m^2-6m+10.$ Ab $a{m}^2+(b-a)m+c$ se match karo jahan $a=1$: humein $b-a=-6$ aur $c=10$ chahiye, toh $b=-5$, $c=10$. $\boxed{x^2{y}^{\prime \prime }-5x{y}^{\prime }+10y=0}$ Check: indicial $m(m-1)-5m+10=m^2-6m+10=0\Rightarrow m=3\pm i$. ✓

(i) Yahan $a=1,b=0,c=-2$. Indicial: $m(m-1)-2=m^2-m-2=(m-2)(m+1)=0\Rightarrow m=2,-1$. $\boxed{y=C_1{x}^2+C_2{x}^{-1}}$ (ii) ODE ko standard form $y^{\prime \prime }+p(x)y^{\prime }+q(x)y=0$ mein likho $x^2$ se divide karke: $y^{\prime \prime }+0\cdot y^{\prime }-\frac{2}{x^2}y=0.$ Coefficient $q(x)=-2/x^2$ blow up karta hai $x=0$ par: original equation ka leading coefficient $x^2$ wahan zero ho jaata hai, toh $x=0$ ek singular point hai. Dhyan do solution $C_2{x}^{-1}$ literally $\pm \infty$ jaata hai jab $x\to 0$ — tum $x=0$ par well-behaved initial condition pose nahi kar sakte. Kyunki $x\cdot p(x)=0$ aur $x^2\cdot q(x)=-2$ finite rehte hain jab $x\to 0$, yeh ek regular singular point hai — exactly woh setting jahan Frobenius method laagoo hoti hai. Actually Cauchy-Euler sabse simple Frobenius case hai: indicial equation yahan wahi Frobenius indicial equation hai, aur power solutions $x^m$ Frobenius series ke leading terms hain jisme koi correction terms ki zarurat nahi.

Distinct real roots $:::y=C_1{x}^{m_1}+C_2{x}^{m_2}$, koi logs nahi. Repeated root $:::y=(C_1+C_2\ln x)x^m$, ek log. Complex roots $\alpha \pm i\beta$ $:::y=x^{\alpha }[C_1\cos (\beta \ln x)+C_2\sin (\beta \ln x)]$. Right side $x^r$ jahan $r$ ek simple indicial root hai $:::y_p$ ko ek extra $\ln x$ factor chahiye. Right side $x^r$ jahan $r$ ek double indicial root hai $:::y_p$ ko $(\ln x{)}^2$ chahiye.