Reduction of order with one solution known

At a Glance

• Frequency: 3 sub-parts across 3 of 13 years (2014, 2019, 2024)
• Priority tier: T3
• Marks (count): 10 (2), 15 (1)
• Average solve time: ~12 min
• Difficulty mix: medium 3
• Section: B | Dominant type: computation

Why This Chapter Matters

Reduction of order is the standard technique when one homogeneous solution is given or guessable — which UPSC has tested in three different years. The 2014 question (15 marks) pairs the substitution with an integrating-factor finish; the 2024 question (10 marks) gives a polynomial second solution from a clean algebra chain; the 2019 question (10 marks) uses a change of independent variable to reduce a messy trig ODE to a constant-coefficient one. All three reduce to the same two-step skeleton: set $y=y_1 v$, find that the $v$-coefficient vanishes, and solve the resulting first-order ODE for $v'$. That skeleton is worth 35 marks across three years.

Minimum Theory

Reduction of order (the substitution). If $y_1(x)$ is one non-trivial solution of the homogeneous ODE $y''+P(x)y'+Q(x)y=0$, set $y=y_1(x)v(x)$. Compute $y'=y_1'v+y_1v'$ and $y''=y_1''v+2y_1'v'+y_1v''$. Substitute into the ODE. Because $y_1$ is a homogeneous solution, the coefficient of $v$ is exactly zero — the equation reduces to a linear first-order ODE in $w=v'$:

$y_1 w' + (2y_1'+Py_1)\,w = R(x) \qquad\text{(or }=0\text{ for the homogeneous case).}$

Solve for $w$ by an integrating factor, then integrate once more to get $v$, giving $y=y_1 v$.

Change of independent variable. If $P(x)$ and $Q(x)$ contain a natural substitution $t=\phi(x)$, convert the ODE to $t$ as the new independent variable. The chain-rule derivatives $dy/dx=Y'(t)\phi'(x)$ and $d^2y/dx^2=Y''(t)(\phi')^2+Y'(t)\phi''(x)$ are substituted; a good choice of $\phi$ makes cross-terms cancel, leaving a constant-coefficient ODE in $t$.

Abel’s identity (Wronskian check). The Wronskian $W[y_1,y_2]=y_1y_2'-y_2y_1'$ satisfies $W=C\exp(-\int P\,dx)$. A second independent solution exists whenever $W\neq0$.

Question Archetypes

reduction-of-order (2 question(s); 2014, 2024)

Recognition Cues

• “Solve the ODE given that $y_1$ is a (homogeneous) solution.”
• “Find the second solution using $u(x)$ as one solution.”
• The ODE is second-order linear with non-constant coefficients; one solution is provided.
• The exponent/polynomial structure of the given solution (e.g., $e^x$, $e^{-x}$, $x$) hints at the substitution.

Solution Template

1. Substitute $y = y_1\,v$. Compute $y'$ and $y''$ using the product rule.
2. Substitute into the ODE. Group terms by $v$, $v'$, $v''$.
3. Verify the $v$-coefficient is zero (since $y_1$ solves the homogeneous equation).
4. Let $w=v'$ to get a first-order ODE: $a(x)w'+b(x)w=R(x)$.
5. Find the integrating factor $\mu = e^{\int(b/a)\,dx}$ and solve for $w$.
6. Integrate $w$ to get $v$; write $y=y_1 v$.
7. Absorb constants and state the general solution (CF + PI for non-homogeneous).

Worked Example

2014 Paper 1, 2014-P1-Q7a (15 marks)

Solve $xy''-2(x+1)y'+(x+2)y=(x-2)e^{2x}$ when $e^x$ is a solution to the corresponding homogeneous equation.

Step 1 — Substitute $y=e^x v$. Using $y'=e^x(v+v')$, $y''=e^x(v+2v'+v'')$:

LHS after substituting and dividing by $e^x$: collect coefficients.

• Coefficient of $v''$: $x$.
• Coefficient of $v'$: $2x - 2(x+1) = -2$.
• Coefficient of $v$: $x - 2(x+1) + (x+2) = 0$. ✓

So the equation reduces to:

$xe^x(v+2v'+v'') - 2(x+1)e^x(v+v') + (x+2)e^xv = (x-2)e^{2x}$

$\Longrightarrow\qquad xv'' - 2v' = (x-2)e^x.$

Step 2 — Let $w=v'$. Now we have a first-order ODE:

$xw' - 2w = (x-2)e^x \qquad\Longrightarrow\qquad w' - \frac{2}{x}w = \left(1-\frac{2}{x}\right)e^x.$

Step 3 — Integrating factor. $\mu = e^{-\int(2/x)\,dx} = x^{-2}$. Multiply:

$\frac{d}{dx}\!\left(\frac{w}{x^2}\right) = \left(\frac{1}{x^2}-\frac{2}{x^3}\right)e^x = \frac{d}{dx}\!\left(\frac{e^x}{x^2}\right).$

Integrate: $\dfrac{w}{x^2} = \dfrac{e^x}{x^2} + C$, so $w = e^x + Cx^2$.

Step 4 — Integrate. $v = \displaystyle\int w\,dx = e^x + \frac{C}{3}x^3 + D$.

Step 5 — General solution. $y = e^x v = e^{2x} + \dfrac{C}{3}x^3 e^x + De^x$. Rename $c_1=D$, $c_2=C/3$:

$\boxed{\;y = c_1 e^x + c_2 x^3 e^x + e^{2x}.\;}$

CF: $c_1 e^x + c_2 x^3 e^x$. Particular integral: $e^{2x}$.

Verify PI: $y=e^{2x}$, $y'=2e^{2x}$, $y''=4e^{2x}$. LHS $= e^{2x}[4x-4(x+1)+(x+2)]=e^{2x}(x-2)$ ✓.

2024 Paper 1, 2024-P1-Q6c-i (10 marks)

Find the second solution of $xy''+(x-1)y'-y=0$ using $u(x)=-e^{-x}$ as one solution.

Step 1 — Substitute $y=-e^{-x}v$. Compute $y'=e^{-x}(v-v')$, $y''=e^{-x}(-v+2v'-v'')$. Substitute:

$x\cdot e^{-x}(-v+2v'-v'')+(x-1)\cdot e^{-x}(v-v')-(-e^{-x})v=0.$

Divide by $e^{-x}$ and expand:

$-xv+2xv'-xv''+xv-xv'-v+v'+v=0.$

Collect: $v$-terms: $0$. $v'$-terms: $(x+1)v'$. $v''$-terms: $-xv''$.

$-xv'' + (x+1)v' = 0.$

Step 2 — Let $w=v'$. Then $-xw'+(x+1)w=0$, so:

$\frac{w'}{w} = \frac{x+1}{x} = 1 + \frac{1}{x}.$

Integrate: $\ln|w| = x + \ln|x| + \text{const}$, so $w = Cxe^x$.

Step 3 — Integrate. $v = C\displaystyle\int xe^x\,dx = C(x-1)e^x + D$ (using integration by parts).

Take $C=1$, $D=0$ (the $D$ term regenerates $u$): $v=(x-1)e^x$.

Step 4 — Second solution. $y_2 = -e^{-x}\cdot(x-1)e^x = -(x-1) = 1-x$.

$\boxed{\;y_2(x) = 1-x.\;}$

Verify: $y_2'=-1$, $y_2''=0$. LHS $= 0+(x-1)(-1)-(1-x)=-(x-1)+(x-1)=0$ ✓.

Common Traps

• The coefficient of $v$ in the substituted equation must vanish — if it does not, there is an algebra error. This is the built-in check.
• After applying the integrating factor in the 2014 question, recognise that $\dfrac{d}{dx}\!\left(\dfrac{e^x}{x^2}\right)=\left(\dfrac{1}{x^2}-\dfrac{2}{x^3}\right)e^x$. Writing out the antiderivative prevents a needless integration by parts.
• When integrating $\int xe^x\,dx$ in the 2024 question, use integration by parts with $u=x$, $dv=e^x\,dx$ to get $(x-1)e^x$.
• The constant of integration in $v$ that reintroduces a multiple of $y_1$ should be dropped (set to zero) when constructing the second independent solution.

change-of-variable (1 question(s); 2019)

Recognition Cues

• Coefficients of $y'$ or $y''$ contain $\sin x$, $\cos x$, $\cot x$, $\tan x$ in a structured way.
• A substitution $t=\phi(x)$ (e.g., $t=-\cos x$, $t=\ln x$) is suggested by the structure.
• After the substitution, cross terms in $Y'$ cancel, leaving a constant-coefficient ODE in $t$.
• The forcing term can be written in terms of $t$ cleanly (e.g., $e^{-\cos x}=e^t$ when $t=-\cos x$).

Solution Template

1. Identify the substitution $t=\phi(x)$. Compute $dt/dx=\phi'(x)$.
2. Transform derivatives. $dy/dx = Y'(t)\phi'$; $d^2y/dx^2 = Y''(t)(\phi')^2 + Y'(t)\phi''$.
3. Substitute. Expand; verify that problematic cross terms in $Y'$ cancel.
4. Divide by the common factor (often $(\phi')^2$) to simplify.
5. Solve the resulting constant-coefficient ODE in $t$ (characteristic equation + particular integral).
6. Back-substitute $t=\phi(x)$ to express the solution in terms of $x$.

Worked Example

2019 Paper 1, 2019-P1-Q6c-i (10 marks)

Solve $y''+(3\sin x-\cot x)y'+2y\sin^2 x = e^{-\cos x}\sin^2 x$.

Step 1 — Choose $t=-\cos x$. Then $dt/dx=\sin x$ and $d^2t/dx^2=\cos x$.

Step 2 — Transform. With $y=Y(t)$:

$\frac{dy}{dx} = Y'(t)\sin x, \qquad \frac{d^2y}{dx^2} = Y''(t)\sin^2 x + Y'(t)\cos x.$

Step 3 — Substitute.

$Y''\sin^2 x + Y'\cos x + (3\sin x-\cot x)\cdot Y'\sin x + 2Y\sin^2 x = e^{-\cos x}\sin^2 x.$

Expand the middle term: $(3\sin x-\cot x)\cdot Y'\sin x = 3Y'\sin^2 x - Y'\cos x$.

The $Y'\cos x$ terms cancel:

$Y''\sin^2 x + 3Y'\sin^2 x + 2Y\sin^2 x = e^{-\cos x}\sin^2 x.$

Step 4 — Divide by $\sin^2 x$. Note $e^{-\cos x}=e^t$:

$Y'' + 3Y' + 2Y = e^t.$

Step 5 — Solve. Characteristic equation: $m^2+3m+2=(m+1)(m+2)=0$, roots $m=-1,-2$.

CF: $C_1 e^{-t}+C_2 e^{-2t}$.

PI: try $Y_p=Ae^t$. Substituting: $(1+3+2)A=1$, so $A=1/6$.

$Y(t) = C_1 e^{-t}+C_2 e^{-2t}+\frac{1}{6}e^t.$

Step 6 — Back-substitute $t=-\cos x$.

$\boxed{\;y = C_1 e^{\cos x}+C_2 e^{2\cos x}+\frac{1}{6}e^{-\cos x}.\;}$

Common Traps

• The substitution is $t=-\cos x$ (negative sign), not $t=\cos x$. With the positive sign, $dt/dx=-\sin x$ and the cancellation of the $Y'\cos x$ cross terms still works, but the sign in $d^2t/dx^2$ flips — either sign can be made to work but the negative convention is cleaner.
• Dividing by $\sin^2 x$ must happen after the cancellation; dividing earlier leaves un-simplified $\cot x$ terms.
• For the PI with forcing $e^t$: since $t=1$ is not a root of $m^2+3m+2$, try $Y_p=Ae^t$ directly. The coefficient is $1/(1+3+2)=1/6$.
• When back-substituting, $e^{-t}=e^{-(-\cos x)}=e^{\cos x}$ and $e^{-2t}=e^{2\cos x}$ — be careful with signs.

Marks-Aware Writing

10-mark questions (2024, 2019): The substitution setup and the vanishing of the $v$-coefficient (or the cross-term cancellation in 2019) each count for 2–3 marks. The integration/solving step is worth 3–4 marks. A verified boxed answer earns the final mark. Showing the $v$-coefficient vanishes is mandatory — it proves the substitution is valid.

15-mark question (2014): The substitution (3 marks), deriving $xw'-2w=(x-2)e^x$ (3 marks), the integrating factor and recognising the antiderivative $e^x/x^2$ (4 marks), integrating $v$ (2 marks), and the final general solution (2 marks), plus verifying the PI (1 mark). Students who apply the integrating factor but cannot identify the exact antiderivative often get stuck; knowing $d/dx\,(e^x/x^2)=(1/x^2-2/x^3)e^x$ avoids a protracted integration by parts.

Practice Set