GATE 2018 EC Q.34: solution and answer

GATE 2018 · ECQ.34 · 2 marks · MCQ

A curve passes through the point $(x = 1, y = 0)$ and satisfies the differential equation $\frac{d y}{d x} = \frac{x ^{2} + y ^{2}}{2 y} + \frac{y}{x} .$ The equation that describes the curve is

(A)
$ln (1 + \frac{y ^{2}}{x ^{2}}) = x - 1$
(B)
$\frac{1}{2} ln (1 + \frac{y ^{2}}{x ^{2}}) = x - 1$
(C)
$ln (1 + \frac{y}{x}) = x - 1$
(D)
$\frac{1}{2} ln (1 + \frac{y}{x}) = x - 1$

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

The curve is $ln (1 + \frac{y ^{2}}{x ^{2}}) = x - 1$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2018 EC (IIT Guwahati)

Substitute $y = v x$ , so $\frac{d y}{d x} = v + x \frac{d v}{d x}$ . The last term of the equation, $\frac{y}{x}$ , is just $v$ .
Rewrite the first term: $\frac{x ^{2} + y ^{2}}{2 y} = \frac{x ^{2} ( 1 + v ^{2} )}{2 v x} = \frac{x ( 1 + v ^{2} )}{2 v}$ .
Substituting gives $v + x \frac{d v}{d x} = \frac{x ( 1 + v ^{2} )}{2 v} + v$ . The $v$ cancels on both sides, leaving $x \frac{d v}{d x} = \frac{x ( 1 + v ^{2} )}{2 v}$ . Divide by $x$ : $\frac{d v}{d x} = \frac{1 + v ^{2}}{2 v}$ .
Separate the variables: $\frac{2 v}{1 + v ^{2}} d v = d x$ . The numerator is the derivative of the denominator, so $\int \frac{f ^{'} ( v )}{f ( v )} d v = ln ∣ f (v) ∣$ with $f = 1 + v^{2}$ gives $ln (1 + v^{2}) = x + C .$ No modulus is needed because $1 + v^{2} > 0$ .
Put $v = y / x$ back: $ln (1 + \frac{y ^{2}}{x ^{2}}) = x + C$ .
Use the point $(1, 0)$ : $ln (1 + 0) = 1 + C$ , so $0 = 1 + C$ and $C = - 1$ . The curve is $ln (1 + \frac{y ^{2}}{x ^{2}}) = x - 1.$
Check: this curve gives $y^{2} = x^{2} (e^{x - 1} - 1)$ . Differentiating, $2 y y^{'} = 2 x (e^{x - 1} - 1) + x^{2} e^{x - 1}$ . From the differential equation, $2 y y^{'} = x^{2} + y^{2} + \frac{2 y ^{2}}{x} = x^{2} + x^{2} (e^{x - 1} - 1) + 2 x (e^{x - 1} - 1) = x^{2} e^{x - 1} + 2 x (e^{x - 1} - 1)$ . The two expressions are the same, so the answer is option (A).

KEY IDEA

The right-hand side depends on $x$ and $y / x$ , so try $y = v x$ . The $y / x$ term cancels the $v$ that comes from $\frac{d y}{d x} = v + x \frac{d v}{d x}$ , and the $x$ factors also cancel. That leaves the separable equation $\frac{2 v}{1 + v ^{2}} d v = d x$ . It integrates directly to $ln (1 + v^{2})$ because the numerator is the derivative of the denominator.

WATCH OUT FOR

Integrating $\frac{2 v}{1 + v ^{2}}$ as $\frac{1}{2} ln (1 + v^{2})$ , which confuses it with $\int \frac{v}{1 + v ^{2}} d v$ . That extra half gives option (B). Options (C) and (D) come from putting back $v$ where $v^{2}$ belongs, i.e. writing $1 + y / x$ where the answer has $1 + y^{2} / x^{2}$ .

From the GATE 2018 EC paper. Every question of this paper · Download the full paper · All 19 Differential equations questions, with notes

GATE 2018 EC Q.34: solution and answer

Answer and solution

More Differential equations questions