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UPSC 2026 Maths Optional Paper 1 Q6a — Step-by-Step Solution 15 marks · Section B
Euler-Cauchy equation · ODEs · asked 9× in 14 yrs · Read the full method →
Question
Solve x 2 d 2 y d x 2 − 3 x d y d x + y = sin ( log x ) + 1 x log x x^2 \dfrac{d^2y}{dx^2} - 3x \dfrac{dy}{dx} + y = \dfrac{\sin(\log x) + 1}{x}\,\log x x 2 d x 2 d 2 y − 3 x d x d y + y = x sin ( log x ) + 1 log x
Technique
Cauchy–Euler (homogeneous linear) equation : put x = e t x=e^t x = e t t = log x t=\log x t = log x θ ≡ d d t \theta\equiv\dfrac{d}{dt} θ ≡ d t d x D → θ xD\to\theta x D → θ x 2 D 2 → θ ( θ − 1 ) x^2D^2\to\theta(\theta-1) x 2 D 2 → θ ( θ − 1 ) e − t ( t sin t + t ) e^{-t}(t\sin t + t) e − t ( t sin t + t ) shift rule 1 F ( θ ) e a t u = e a t 1 F ( θ + a ) u \dfrac{1}{F(\theta)}e^{at}u = e^{at}\dfrac{1}{F(\theta+a)}u F ( θ ) 1 e a t u = e a t F ( θ + a ) 1 u t sin t t\sin t t sin t t sin t = ℑ ( t e i t ) t\sin t=\Im(te^{it}) t sin t = ℑ ( t e i t )
Solution
The equation is posed on x > 0 x>0 x > 0 log x \log x log x
Step 1 — Reduce to constant coefficients.
Put x = e t x=e^{t} x = e t t = log x t=\log x t = log x θ = d d t \theta=\dfrac{d}{dt} θ = d t d
x d y d x = θ y , x 2 d 2 y d x 2 = θ ( θ − 1 ) y . x\frac{dy}{dx}=\theta y,\qquad x^{2}\frac{d^{2}y}{dx^{2}}=\theta(\theta-1)y . x d x d y = θ y , x 2 d x 2 d 2 y = θ ( θ − 1 ) y . The left side becomes
θ ( θ − 1 ) y − 3 θ y + y = ( θ 2 − 4 θ + 1 ) y . \theta(\theta-1)y-3\theta y+y=\bigl(\theta^{2}-4\theta+1\bigr)y . θ ( θ − 1 ) y − 3 θ y + y = ( θ 2 − 4 θ + 1 ) y . The right side, with x = e t x=e^{t} x = e t log x = t \log x=t log x = t
sin t + 1 e t ⋅ t = e − t ( t sin t + t ) . \frac{\sin t+1}{e^{t}}\cdot t=e^{-t}\bigl(t\sin t+t\bigr). e t sin t + 1 ⋅ t = e − t ( t sin t + t ) . So the transformed equation is
( θ 2 − 4 θ + 1 ) y = e − t ( t sin t + t ) . (1) \boxed{\;\bigl(\theta^{2}-4\theta+1\bigr)y=e^{-t}\bigl(t\sin t+t\bigr).\;}\tag{1} ( θ 2 − 4 θ + 1 ) y = e − t ( t sin t + t ) . ( 1 ) Step 2 — Complementary function.
Auxiliary equation m 2 − 4 m + 1 = 0 ⇒ m = 4 ± 16 − 4 2 = 2 ± 3 m^{2}-4m+1=0\Rightarrow m=\dfrac{4\pm\sqrt{16-4}}{2}=2\pm\sqrt3 m 2 − 4 m + 1 = 0 ⇒ m = 2 4 ± 16 − 4 = 2 ± 3
y c = c 1 e ( 2 + 3 ) t + c 2 e ( 2 − 3 ) t = c 1 x 2 + 3 + c 2 x 2 − 3 . y_{c}=c_{1}e^{(2+\sqrt3)t}+c_{2}e^{(2-\sqrt3)t}=c_{1}x^{2+\sqrt3}+c_{2}x^{2-\sqrt3}. y c = c 1 e ( 2 + 3 ) t + c 2 e ( 2 − 3 ) t = c 1 x 2 + 3 + c 2 x 2 − 3 . Step 3 — Strip the exponential e − t e^{-t} e − t
Write F ( θ ) = θ 2 − 4 θ + 1 F(\theta)=\theta^{2}-4\theta+1 F ( θ ) = θ 2 − 4 θ + 1 a = − 1 a=-1 a = − 1
y p = 1 F ( θ ) e − t ( t sin t + t ) = e − t 1 F ( θ − 1 ) ( t sin t + t ) , y_{p}=\frac{1}{F(\theta)}\,e^{-t}\bigl(t\sin t+t\bigr)=e^{-t}\,\frac{1}{F(\theta-1)}\bigl(t\sin t+t\bigr), y p = F ( θ ) 1 e − t ( t sin t + t ) = e − t F ( θ − 1 ) 1 ( t sin t + t ) , and
F ( θ − 1 ) = ( θ − 1 ) 2 − 4 ( θ − 1 ) + 1 = θ 2 − 6 θ + 6. F(\theta-1)=(\theta-1)^{2}-4(\theta-1)+1=\theta^{2}-6\theta+6 . F ( θ − 1 ) = ( θ − 1 ) 2 − 4 ( θ − 1 ) + 1 = θ 2 − 6 θ + 6. So, with Φ ( θ ) ≡ θ 2 − 6 θ + 6 \Phi(\theta)\equiv\theta^{2}-6\theta+6 Φ ( θ ) ≡ θ 2 − 6 θ + 6
y p = e − t [ 1 Φ ( θ ) t ⏟ P 1 + 1 Φ ( θ ) t sin t ⏟ P 2 ] . (2) y_{p}=e^{-t}\left[\underbrace{\frac{1}{\Phi(\theta)}\,t}_{P_{1}}+\underbrace{\frac{1}{\Phi(\theta)}\,t\sin t}_{P_{2}}\right].\tag{2} y p = e − t P 1 Φ ( θ ) 1 t + P 2 Φ ( θ ) 1 t sin t . ( 2 ) Note Φ ( 0 ) = 6 ≠ 0 \Phi(0)=6\neq0 Φ ( 0 ) = 6 = 0 Φ ( ± i ) ≠ 0 \Phi(\pm i)\neq0 Φ ( ± i ) = 0 t t t
Step 4 — The easy piece P 1 P_{1} P 1
Expand Φ ( θ ) − 1 \Phi(\theta)^{-1} Φ ( θ ) − 1 θ \theta θ θ 1 \theta^{1} θ 1 θ 2 t = 0 \theta^{2}t=0 θ 2 t = 0
1 6 − 6 θ + θ 2 = 1 6 [ 1 − 6 θ − θ 2 6 ] − 1 = 1 6 [ 1 + θ + ⋯ ] . \frac{1}{6-6\theta+\theta^{2}}=\frac{1}{6}\left[1-\frac{6\theta-\theta^{2}}{6}\right]^{-1}=\frac{1}{6}\Bigl[1+\theta+\cdots\Bigr]. 6 − 6 θ + θ 2 1 = 6 1 [ 1 − 6 6 θ − θ 2 ] − 1 = 6 1 [ 1 + θ + ⋯ ] . Hence
P 1 = 1 6 ( t + 1 ) . P_{1}=\frac{1}{6}(t+1). P 1 = 6 1 ( t + 1 ) . (Check on the spot: with Y = 1 6 ( t + 1 ) Y=\tfrac{1}{6}(t+1) Y = 6 1 ( t + 1 ) θ 2 Y − 6 θ Y + 6 Y = 0 − 1 + ( t + 1 ) = t \ \theta^{2}Y-6\theta Y+6Y=0-1+(t+1)=t θ 2 Y − 6 θ Y + 6 Y = 0 − 1 + ( t + 1 ) = t
Step 5 — The hard piece P 2 P_{2} P 2 1 Φ ( θ ) t sin t \dfrac{1}{\Phi(\theta)}\,t\sin t Φ ( θ ) 1 t sin t
Write t sin t = ℑ ( t e i t ) t\sin t=\Im\!\left(te^{it}\right) t sin t = ℑ ( t e i t ) a = i a=i a = i
P 2 = ℑ [ 1 Φ ( θ ) t e i t ] = ℑ [ e i t 1 Φ ( θ + i ) t ] . P_{2}=\Im\left[\frac{1}{\Phi(\theta)}\,t\,e^{it}\right]=\Im\left[e^{it}\,\frac{1}{\Phi(\theta+i)}\,t\right]. P 2 = ℑ [ Φ ( θ ) 1 t e i t ] = ℑ [ e i t Φ ( θ + i ) 1 t ] . Compute the shifted operator:
Φ ( θ + i ) = ( θ + i ) 2 − 6 ( θ + i ) + 6 = θ 2 + ( 2 i − 6 ) θ + ( 5 − 6 i ) . \Phi(\theta+i)=(\theta+i)^{2}-6(\theta+i)+6=\theta^{2}+(2i-6)\theta+(5-6i). Φ ( θ + i ) = ( θ + i ) 2 − 6 ( θ + i ) + 6 = θ 2 + ( 2 i − 6 ) θ + ( 5 − 6 i ) . Since θ t = 1 \theta t=1 θ t = 1 θ 2 t = 0 \theta^{2}t=0 θ 2 t = 0
1 Φ ( θ + i ) t = 1 5 − 6 i [ 1 + ( 2 i − 6 ) θ + θ 2 5 − 6 i ] − 1 t = 1 5 − 6 i [ t − 2 i − 6 5 − 6 i ] = 1 5 − 6 i [ t + 6 − 2 i 5 − 6 i ] . \frac{1}{\Phi(\theta+i)}\,t=\frac{1}{5-6i}\left[1+\frac{(2i-6)\theta+\theta^{2}}{5-6i}\right]^{-1}t
=\frac{1}{5-6i}\left[t-\frac{2i-6}{5-6i}\right]=\frac{1}{5-6i}\left[t+\frac{6-2i}{5-6i}\right]. Φ ( θ + i ) 1 t = 5 − 6 i 1 [ 1 + 5 − 6 i ( 2 i − 6 ) θ + θ 2 ] − 1 t = 5 − 6 i 1 [ t − 5 − 6 i 2 i − 6 ] = 5 − 6 i 1 [ t + 5 − 6 i 6 − 2 i ] . Rationalise, using 1 5 − 6 i = 5 + 6 i 61 \dfrac{1}{5-6i}=\dfrac{5+6i}{61} 5 − 6 i 1 = 61 5 + 6 i
6 − 2 i 5 − 6 i = ( 6 − 2 i ) ( 5 + 6 i ) 61 = 30 + 36 i − 10 i + 12 61 = 42 + 26 i 61 , \frac{6-2i}{5-6i}=\frac{(6-2i)(5+6i)}{61}=\frac{30+36i-10i+12}{61}=\frac{42+26i}{61}, 5 − 6 i 6 − 2 i = 61 ( 6 − 2 i ) ( 5 + 6 i ) = 61 30 + 36 i − 10 i + 12 = 61 42 + 26 i , 1 Φ ( θ + i ) t = 5 + 6 i 61 [ t + 42 + 26 i 61 ] = ( 5 + 6 i ) t 61 + ( 5 + 6 i ) ( 42 + 26 i ) 3721 , \frac{1}{\Phi(\theta+i)}t=\frac{5+6i}{61}\left[t+\frac{42+26i}{61}\right]=\frac{(5+6i)t}{61}+\frac{(5+6i)(42+26i)}{3721}, Φ ( θ + i ) 1 t = 61 5 + 6 i [ t + 61 42 + 26 i ] = 61 ( 5 + 6 i ) t + 3721 ( 5 + 6 i ) ( 42 + 26 i ) , and ( 5 + 6 i ) ( 42 + 26 i ) = 210 + 130 i + 252 i − 156 = 54 + 382 i (5+6i)(42+26i)=210+130i+252i-156=54+382i ( 5 + 6 i ) ( 42 + 26 i ) = 210 + 130 i + 252 i − 156 = 54 + 382 i 61 2 = 3721 61^{2}=3721 6 1 2 = 3721
1 Φ ( θ + i ) t = ( 5 t 61 + 54 3721 ) ⏟ A + i ( 6 t 61 + 382 3721 ) ⏟ B . \frac{1}{\Phi(\theta+i)}t=\underbrace{\left(\frac{5t}{61}+\frac{54}{3721}\right)}_{A}+i\underbrace{\left(\frac{6t}{61}+\frac{382}{3721}\right)}_{B}. Φ ( θ + i ) 1 t = A ( 61 5 t + 3721 54 ) + i B ( 61 6 t + 3721 382 ) . Finally, ℑ [ ( cos t + i sin t ) ( A + i B ) ] = A sin t + B cos t \Im\bigl[(\cos t+i\sin t)(A+iB)\bigr]=A\sin t+B\cos t ℑ [ ( cos t + i sin t ) ( A + i B ) ] = A sin t + B cos t
P 2 = ( 5 t 61 + 54 3721 ) sin t + ( 6 t 61 + 382 3721 ) cos t = ( 305 t + 54 ) sin t + ( 366 t + 382 ) cos t 3721 . P_{2}=\left(\frac{5t}{61}+\frac{54}{3721}\right)\sin t+\left(\frac{6t}{61}+\frac{382}{3721}\right)\cos t
=\frac{(305t+54)\sin t+(366t+382)\cos t}{3721}. P 2 = ( 61 5 t + 3721 54 ) sin t + ( 61 6 t + 3721 382 ) cos t = 3721 ( 305 t + 54 ) sin t + ( 366 t + 382 ) cos t . Step 6 — Assemble and return to x x x
From (2),
y p = e − t [ t + 1 6 + ( 305 t + 54 ) sin t + ( 366 t + 382 ) cos t 3721 ] . y_{p}=e^{-t}\left[\frac{t+1}{6}+\frac{(305t+54)\sin t+(366t+382)\cos t}{3721}\right]. y p = e − t [ 6 t + 1 + 3721 ( 305 t + 54 ) sin t + ( 366 t + 382 ) cos t ] . Substituting back e − t = 1 x e^{-t}=\dfrac1x e − t = x 1 t = log x t=\log x t = log x
y = c 1 x 2 + 3 + c 2 x 2 − 3 + 1 x [ 1 + log x 6 + ( 305 log x + 54 ) sin ( log x ) + ( 366 log x + 382 ) cos ( log x ) 3721 ] . y=c_{1}x^{2+\sqrt3}+c_{2}x^{2-\sqrt3}+\frac{1}{x}\left[\frac{1+\log x}{6}+\frac{(305\log x+54)\sin(\log x)+(366\log x+382)\cos(\log x)}{3721}\right]. y = c 1 x 2 + 3 + c 2 x 2 − 3 + x 1 [ 6 1 + log x + 3721 ( 305 log x + 54 ) sin ( log x ) + ( 366 log x + 382 ) cos ( log x ) ] . Answer
y = c 1 x 2 + 3 + c 2 x 2 − 3 + 1 + log x 6 x + ( 305 log x + 54 ) sin ( log x ) + ( 366 log x + 382 ) cos ( log x ) 3721 x \boxed{\;y=c_{1}x^{2+\sqrt3}+c_{2}x^{2-\sqrt3}+\frac{1+\log x}{6x}+\frac{(305\log x+54)\sin(\log x)+(366\log x+382)\cos(\log x)}{3721\,x}\;} y = c 1 x 2 + 3 + c 2 x 2 − 3 + 6 x 1 + log x + 3721 x ( 305 log x + 54 ) sin ( log x ) + ( 366 log x + 382 ) cos ( log x )