UPSC 2026 Maths Optional Paper 1 Q5a — Step-by-Step Solution

10 marks · Section B

Linear first-order · ODEs · asked 6× in 14 yrs · Read the full method →

Question

If $y_1$ and $y_2$ are two solutions of $\dfrac{dy}{dx} + P(x)y = Q(x)$ , and $y_2 = y_1 u$ , then show that

u = 1 + a\,e^{-\int \left(\frac{Q}{y_1}\right)dx},

where $a$ is a non-zero constant.

Technique

Substitute $y_2=y_1u$ into the linear equation and use the fact that $y_1$ already satisfies it to annihilate the $P$ -terms; what survives is a separable first-order equation for $u$ alone. Integrate, and read off why the constant cannot be zero.

Solution

Step 1 — State the hypotheses precisely (and the standing assumption).

Both functions satisfy the same linear equation on a common interval $I$ :

\frac{dy_1}{dx} + P(x)\,y_1 = Q(x), \tag{1}

\frac{dy_2}{dx} + P(x)\,y_2 = Q(x). \tag{2}

For the substitution $y_2=y_1u$ to define $u$ at all we need $y_1\neq 0$ on $I$ ; and for $y_2$ to be a second (i.e. different) solution we need $y_2\not\equiv y_1$ , so $u\not\equiv 1$ . Both are taken throughout.

Step 2 — Put $y_2=y_1u$ into equation (2).

By the product rule, $\dfrac{dy_2}{dx} = y_1\dfrac{du}{dx} + u\dfrac{dy_1}{dx}$ . Substituting in (2):

y_1\frac{du}{dx} + u\frac{dy_1}{dx} + P\,y_1u = Q,

and grouping the two terms carrying $u$ :

y_1\frac{du}{dx} + u\underbrace{\left(\frac{dy_1}{dx} + P\,y_1\right)}_{=\;Q\ \text{by (1)}} = Q.

This is the load-bearing step: the bracket is exactly the left side of (1), so it equals $Q$ . Note that $P$ has now disappeared completely. Hence

y_1\frac{du}{dx} + uQ = Q \qquad\Longrightarrow\qquad y_1\frac{du}{dx} = Q\,(1-u). \tag{3}

Step 3 — Separate and integrate.

Equation (3) is separable. Since $u\not\equiv1$ we may divide by $1-u$ (on any subinterval where $1-u\neq0$ ; by uniqueness for the linear equation, $u=1$ at one point would force $y_2\equiv y_1$ , so in fact $1-u$ never vanishes):

\frac{du}{1-u} = \frac{Q}{y_1}\,dx.

Integrating both sides,

-\ln|1-u| = \int\frac{Q}{y_1}\,dx + c,

|1-u| = e^{-c}\,e^{-\int (Q/y_1)\,dx}\qquad\Longrightarrow\qquad 1-u = A\,e^{-\int (Q/y_1)\,dx},

where $A=\pm e^{-c}$ is an arbitrary non-zero constant (the sign is absorbed on removing the modulus).

Therefore

u = 1 - A\,e^{-\int (Q/y_1)\,dx} = 1 + a\,e^{-\int (Q/y_1)\,dx},\qquad a:=-A .

Step 4 — Why $a\neq0$ .

If $a=0$ the formula gives $u\equiv1$ , hence $y_2=y_1u=y_1$ . But $y_1$ and $y_2$ were taken to be two solutions, i.e. distinct ones; a repeat of $y_1$ carries no new information and the substitution $y_2=y_1u$ would be vacuous. Hence $a\neq0$ . (Equivalently: $a=0$ is exactly the degenerate case excluded by $u\not\equiv1$ in Step 2, which is what allowed the division by $1-u$ .)

Step 5 — Consistency check (worth one line in the exam).

Subtracting (1) from (2), $w:=y_2-y_1$ satisfies the homogeneous equation $w'+Pw=0$ , so $w=Ce^{-\int P\,dx}$ . Our answer gives

w = y_1(u-1) = a\,y_1\,e^{-\int (Q/y_1)\,dx}.

The two agree because, dividing (1) by $y_1$ ,

\frac{y_1'}{y_1} - \frac{Q}{y_1} = -P \quad\Longrightarrow\quad \frac{d}{dx}\!\left[\ln y_1 - \int\frac{Q}{y_1}dx\right] = -P,

i.e. $y_1e^{-\int (Q/y_1)dx}=e^{\,\ln y_1-\int (Q/y_1)dx}$ is precisely $e^{-\int P\,dx}$ up to a constant. The result is therefore internally consistent with the standard structure “general solution = particular + homogeneous”.

Answer

\boxed{\;u = \frac{y_2}{y_1} = 1 + a\,e^{-\int \left(\frac{Q}{y_1}\right)dx},\qquad a\neq0\ \ \text{(else }y_2\equiv y_1).\;}