GATE 2020 EC Q.27: solution and answer

GATE 2020 · ECQ.27 · 2 marks · MCQ

Which one of the following options contains two solutions of the differential equation $\frac{d y}{d x} = (y - 1) x$ ?

(A)
$ln ∣ y - 1∣ = 0.5 x^{2} + C$ and $y = 1$
(B)
$ln ∣ y - 1∣ = 2 x^{2} + C$ and $y = 1$
(C)
$ln ∣ y - 1∣ = 0.5 x^{2} + C$ and $y = - 1$
(D)
$ln ∣ y - 1∣ = 2 x^{2} + C$ and $y = - 1$

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Answer and solution

ANSWER

Option (A): $ln ∣ y - 1∣ = 0.5 x^{2} + C$ and $y = 1$ are both solutions of $\frac{d y}{d x} = (y - 1) x$ .

Official answer: ASource: Official answer key, GATE 2020 EC (IIT Delhi)

The equation $\frac{d y}{d x} = (y - 1) x$ is variables separable: the right side is a function of $y$ multiplied by a function of $x$ .
For $y \neq = 1$ , separate the variables: $\frac{d y}{y - 1} = x d x$
Integrate both sides: $\int \frac{d y}{y - 1} = \int x d x \Rightarrow ln ∣ y - 1∣ = \frac{x ^{2}}{2} + C = 0.5 x^{2} + C$ The coefficient of $x^{2}$ is $0.5$ , not $2$ , which rules out (B) and (D). (Check: differentiating $ln ∣ y - 1∣ = 0.5 x^{2} + C$ gives $\frac{1}{y - 1} \frac{d y}{d x} = x$ , i.e. $\frac{d y}{d x} = (y - 1) x$ .)
Constant (equilibrium) solution: dividing by $y - 1$ assumed $y \neq = 1$ , so $y = 1$ must be checked separately. LHS $= \frac{d}{d x} (1) = 0$ and RHS $= (1 - 1) x = 0$ , so $y = 1$ satisfies the equation for every $x$ . It is not contained in $ln ∣ y - 1∣ = 0.5 x^{2} + C$ for any finite $C$ , which is why it is listed as a second solution.
Check $y = - 1$ : LHS $= 0$ but RHS $= (- 1 - 1) x = - 2 x$ , which is not zero for $x \neq = 0$ . So $y = - 1$ is not a solution, which rules out (C).
The two solutions are $ln ∣ y - 1∣ = 0.5 x^{2} + C$ and $y = 1$ : option (A).

KEY IDEA

Separate the variables to get the general family $ln ∣ y - 1∣ = \frac{1}{2} x^{2} + C$ . Then remember that dividing by $y - 1$ threw away the case $y = 1$ . Substituting it back shows that the constant $y = 1$ is also a solution.

WATCH OUT FOR

There are two traps. One is getting the coefficient wrong when integrating $x d x$ (it gives $x^{2} /2$ , not $2 x^{2}$ ). The other is picking $y = - 1$ as the constant solution. The constant solution is the root of the $y$ -factor, $y - 1 = 0$ , so it is $y = 1$ . Put any candidate constant back into the equation: $y = - 1$ gives $0 = - 2 x$ , which fails.

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GATE 2020 EC Q.27: solution and answer

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