Euler-Cauchy equation

At a Glance

• Frequency: 8 sub-parts across 8 of 13 years (2013, 2014, 2015, 2016, 2018, 2020, 2022, 2025)
• Priority tier: T1
• Marks (count): 10 (3), 13 (2), 15 (2), 20 (1)
• Average solve time: ~16 min
• Difficulty mix: medium 5, hard 3
• Section: B | Dominant type: computation

Why This Chapter Matters

The Cauchy-Euler (equidimensional) equation appears in 8 of the last 13 years — no year with a known UPSC result is missing it entirely. The method is a single substitution ($x=e^t$) that converts a variable-coefficient ODE into a constant-coefficient one, then the full machinery of operator methods applies. Three variants appear: standard $x^n$, shifted $(a+bx)^n$, and order-reduction when the $y$ term is absent. Mastering one set of operator identities covers all of them.

Minimum Theory

Cauchy-Euler operator identities. Under $x=e^t$, $t=\ln x$, $D=d/dt$: $x\frac{dy}{dx}=Dy,\qquad x^2\frac{d^2y}{dx^2}=D(D-1)y,\qquad x^3\frac{d^3y}{dx^3}=D(D-1)(D-2)y.$ In general: $x^n y^{(n)}=D(D-1)\cdots(D-n+1)y$. After substitution, the ODE has constant coefficients in $t$.

Generalised Cauchy-Euler. If the equation has $(ax+b)$ instead of $x$: substitute $ax+b=e^t$ (equivalently, $t=\ln(ax+b)$). The operators become $(ax+b)y'=aDy$ and $(ax+b)^2y''=a^2D(D-1)y$ — note the factors of $a$.

Order reduction. If the $y$ term (zero-th order) is absent and the lowest term is $xy'$, substitute $p=y'$ to convert to a second-order Cauchy-Euler in $p$, then integrate once.

Resonance in the PI. After the $x=e^t$ substitution, if the RHS is $e^{mt}$ and $m$ is a root of the auxiliary equation: multiply the trial PI by $t$. If $m$ is a double root: multiply by $t^2$. This is the standard resonance rule.

Operator formula for PI. For $y_p=\frac{1}{f(D)}e^{at}$: if $f(a)\ne0$, then $y_p=e^{at}/f(a)$. For $\cos bt$ or $\sin bt$: replace $D^2\to-b^2$ (if no cancellation). For resonance: $y_p=t\cdot e^{at}/f'(a)$.

Question Archetypes

One procedure covers all questions.

cauchy-euler (8 question(s); 2013, 2014, 2015, 2016, 2018, 2020, 2022, 2025)

Recognition Cues

• Powers of $x$ match the order of differentiation: $x^2y''+xy'+y=\ldots$ (2nd order), $x^3y'''+\ldots$ (3rd order).
• Or shifted: $(1+x)^2y''+(1+x)y'$, $(x+1)^2y''-4(x+1)y'$, $(3x+2)^2y''+5(3x+2)y'$.
• The RHS may be $\ln x\sin(\ln x)$, $\cos(\log x)$, $x^2$, or a combination.

Solution Template

1. If the equation has $(ax+b)$ instead of $x$: substitute $u=ax+b$, noting $dy/dx=a\,dy/du$, $d^2y/dx^2=a^2\,d^2y/du^2$.
2. Substitute $u=e^t$ (or $x=e^t$ for the standard case). Write $uD^k_u=D(D-1)\cdots(D-k+1)$ where $D=d/dt$.
3. Expand the operator polynomial; find CF by solving the auxiliary equation.
4. Find PI using operator methods. Check for resonance (RHS = CF mode): multiply by $t$.
5. Back-substitute: $t=\ln x$ (or $t=\ln(ax+b)$); $e^t=x$ (or $e^t=ax+b$).

Worked Example(s)

2014 Paper 1, 2014-P1-Q6b (20 marks)

Solve $x^3y'''+3x^2y''+xy'+8y=65\cos(\log x)$.

Substitute. $x=e^t$: $x^3y'''=D(D-1)(D-2)y$; $3x^2y''=3D(D-1)y$; $xy'=Dy$. Operator sum: $D^3-3D^2+2D+3D^2-3D+D+8=D^3+8.$

CF. $D^3+8=0$: roots $D=-2,\;1\pm i\sqrt3$. $y_c=c_1e^{-2t}+e^t(c_2\cos\sqrt3\,t+c_3\sin\sqrt3\,t).$

PI. $\frac{65\cos t}{D^3+8}$. Replace $D^2\to-1$, so $D^3\to-D$: $D^3+8\to8-D$. $y_p=\frac{65(8\cos t+D\cos t)}{(8-D)(8+D)}=\frac{65(8\cos t-\sin t)}{64+1}=8\cos t-\sin t.$

General solution (back-substitute $t=\log x$): $\boxed{\;y=\frac{c_1}{x^2}+x\big(c_2\cos\sqrt3\log x+c_3\sin\sqrt3\log x\big)+8\cos(\log x)-\sin(\log x).\;}$

2013 Paper 1, 2013-P1-Q6c (15 marks)

Solve $x^2y''+xy'+y=\ln x\sin(\ln x)$.

Substitute. $D^2+1=0$: CF $y_c=C_1\cos t+C_2\sin t$ (resonant!). RHS $=t\sin t=\operatorname{Im}(te^{it})$.

PI via complex method. Substitute $y=e^{it}u$: $u''+2iu'=t$. Let $v=u'$: IF $e^{2it}$ gives $v=-it/2+1/4$. $y_p=\operatorname{Im}\!\left[e^{it}\!\left(\tfrac t4-\tfrac{it^2}4\right)\right]=\frac{t\sin t-t^2\cos t}{4}.$

Back-substitute: $\boxed{\;y=C_1\cos(\ln x)+C_2\sin(\ln x)+\frac{\ln x\sin(\ln x)-(\ln x)^2\cos(\ln x)}{4}.\;}$

2025 Paper 1, 2025-P1-Q7c-ii (10 marks)

Solve $x^3y'''+3x^2y''+xy'+y=x\log x$.

Operator. $D(D-1)(D-2)+3D(D-1)+D+1=D^3+1=(D+1)(D^2-D+1)$.

CF. Roots $D=-1,\frac{1\pm i\sqrt3}{2}$: $y_c=\frac{C_1}{x}+\sqrt{x}\!\left[C_2\cos\!\left(\tfrac{\sqrt3}{2}\ln x\right)+C_3\sin\!\left(\tfrac{\sqrt3}{2}\ln x\right)\right].$

PI. RHS $=te^t$. Shift: $\frac{1}{D^3+1}e^t\cdot t=e^t\frac{1}{(D+1)^3+1}t$. Expand $(D+1)^3+1=D^3+3D^2+3D+2$; keep first order: $\approx 2+3D$: $y_p=e^t\cdot\tfrac12\!\left(1-\tfrac32 D+\ldots\right)t=e^t\cdot\tfrac12\!\left(t-\tfrac32\right)=\frac{x\log x}{2}-\frac{3x}{4}.$

$\boxed{\;y=\frac{C_1}{x}+\sqrt x\!\left[C_2\cos\!\left(\tfrac{\sqrt3}{2}\ln x\right)+C_3\sin\!\left(\tfrac{\sqrt3}{2}\ln x\right)\right]+\frac{x\ln x}{2}-\frac{3x}{4}.\;}$

2015 Paper 1, 2015-P1-Q8d (13 marks) — Fourth-order with double resonance

Solve $x^4y^{(4)}+6x^3y'''+4x^2y''-2xy'-4y=x^2+2\cos(\log x)$.

Operator. $D^4-3D^2-4=(D^2-4)(D^2+1)=(D-2)(D+2)(D^2+1)$. Roots: $\pm2,\pm i$.

Both $e^{2t}$ and $\cos t$ resonate with the CF.

PI for $e^{2t}$ (resonance, $f'(2)=20$): $y_{p1}=te^{2t}/20$.

PI for $2\cos t$ (resonance with $D^2+1$, solve $z''+z=2\cos t\Rightarrow z=t\sin t$, then $(D^2-4)y_{p2}=t\sin t$): $y_{p2}=-t\sin t/5$.

$\boxed{\;y=c_1x^2+\frac{c_2}{x^2}+c_3\cos(\ln x)+c_4\sin(\ln x)+\frac{x^2\ln x}{20}-\frac{\ln x\sin(\ln x)}{5}.\;}$

2016 Paper 1, 2016-P1-Q6c (15 marks) — Order reduction ($y$ absent)

Solve $x^2y'''-4xy''+6y'=4$.

Reduce. No $y$ term — let $p=y'$. Then $x^2p''-4xp'+6p=4$ (Cauchy-Euler in $p$).

Indicial equation (from $x=e^t$): $D^2-5D+6=(D-2)(D-3)=0$; roots $m=2,3$.

CF: $p_c=Ax^2+Bx^3$. PI: $p=2/3$ (constant trial). So $p=Ax^2+Bx^3+2/3$.

Integrate: $y=c_1+c_2x^3+c_3x^4+\frac23x$.

$\boxed{\;y=c_1+c_2x^3+c_3x^4+\frac{2x}{3}.\;}$

2018 Paper 1, 2018-P1-Q7a (13 marks) — Shifted base $(1+x)$

Solve $(1+x)^2y''+(1+x)y'+y=4\cos(\log(1+x))$.

Substitute $t=\ln(1+x)$, $D=d/dt$. Operator: $D(D-1)+D+1=D^2+1$. CF $=C_1\cos t+C_2\sin t$.

RHS $=4\cos t$ — resonant. PI: $\frac{1}{D^2+1}\cos t=\frac{t}{2}\sin t$, so $y_p=2t\sin t$.

$\boxed{\;y=C_1\cos\!\big(\ln(1+x)\big)+C_2\sin\!\big(\ln(1+x)\big)+2\ln(1+x)\sin\!\big(\ln(1+x)\big).\;}$

2020 Paper 1, 2020-P1-Q8a-i (10 marks) — Shifted base $(x+1)$

Solve $(x+1)^2y''-4(x+1)y'+6y=6(x+1)^2+\sin\log(x+1)$.

$t=\ln(x+1)$; operator $D^2-5D+6=(D-2)(D-3)$. CF $=C_1e^{2t}+C_2e^{3t}$.

PI for $6e^{2t}$ (resonance, $f'(2)=-1$): $y_{p1}=-6te^{2t}$.

PI for $\sin t$: $D^2\to-1$ gives $f(D)\to5-5D$; $y_{p2}=\frac{\sin t}{5(1-D)}=\frac{1+D}{10}\sin t=\frac{\sin t+\cos t}{10}$.

$\boxed{\;y=C_1(x+1)^2+C_2(x+1)^3-6(x+1)^2\ln(x+1)+\tfrac{1}{10}\big[\sin\ln(x+1)+\cos\ln(x+1)\big].\;}$

2022 Paper 1, 2022-P1-Q8a-ii (10 marks) — Shifted base $(3x+2)$

Solve $(3x+2)^2y''+5(3x+2)y'-3y=x^2+x+1$.

Substitute $u=3x+2$: $dy/dx=3\,dy/du$, $d^2y/dx^2=9\,d^2y/du^2$. Then $u=e^t$; operator $3D^2+2D-1=(3D-1)(D+1)$. Roots $D=1/3,-1$.

CF: $C_1u^{1/3}+C_2/u=C_1(3x+2)^{1/3}+C_2/(3x+2)$.

Express RHS in $u$: $x=(u-2)/3$, so $x^2+x+1=(u^2-u+7)/9$; the $t$-form is $(e^{2t}-e^t+7)/27$. PI for each exponential: $e^{2t}/(27\cdot15)$, $-e^t/(27\cdot4)$, $7/(27\cdot(-1))$.

$\boxed{\;y=C_1(3x+2)^{1/3}+\frac{C_2}{3x+2}+\frac{(3x+2)^2}{405}-\frac{3x+2}{108}-\frac{7}{27}.\;}$

Common Traps

• Operator expansion. For $x^3y'''$: $D(D-1)(D-2)=D^3-3D^2+2D$ — three terms, not just $D^3$. Missing the lower-order terms changes the auxiliary equation.
• Resonance check. After writing the constant-coefficient ODE, compare each RHS term to the CF modes. $\cos t$ resonates with $D^2+1=0$; $e^{2t}$ resonates with $D=2$. In each case, multiply the trial PI by $t$ (or use $y_p=te^{at}/f'(a)$).
• Shifted variables. For $(ax+b)^ky^{(k)}$: factors of $a$ accumulate in the derivative conversion. $(3x+2)y'=3Dy$ but $(3x+2)^2y''=9D(D-1)y$. Factor of $9$, not $1$.
• Order reduction. If there is no $y$ term in the ODE, do NOT try to write the Cauchy-Euler operators directly on $y$ for a cubic. Substitute $p=y'$ first; the resulting equation is second-order.
• The third arbitrary constant in the 2016 problem enters via the final integration (not from the auxiliary equation) — easy to forget.
• Back-substitution. $e^{2t}=x^2$, $te^{2t}=x^2\ln x$, $t\cos t=\ln x\cos(\ln x)$. Write every term in terms of $\ln x$ and powers of $x$ (or $ax+b$).

Marks-Aware Writing

10-mark questions (2020, 2022, 2025): State the substitution; write the constant-coefficient ODE; state roots; write CF; compute PI (show the operator calculation); state general solution with back-substitution. Four to five steps; don’t expand operator products in full — just state the result.

13-mark questions (2015-Q8d, 2018): Full operator expansion showing how the coefficients combine; roots; CF; PI with resonance handling shown (state $f(m)=0$ and use $f'$ formula); general solution in $x$.

15-mark questions (2013, 2014, 2016, 2025): Same as 13-mark plus one of: complex-method PI (2013), third-order CF with cube roots (2014), order-reduction via $p=y'$ (2016), shift operator for PI (2025).

20-mark questions (2014): Full operator expansion for a third-order equation, all three cube roots computed, two-step PI (operator method including the $i$ computation), full back-substitution.

Practice Set