SATHEE: Differential Equations 3 Question 6

Differential Equations 3 Question 6

6.

A solution curve of the differential equation $(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0, x > 0$ , passes through the point $(1, 3)$ . Then, the solution curve

(2016 Adv.)

(a) intersects $y = x + 2$ exactly at one point

(b) intersects $y = x + 2$ exactly at two points

(d) does not intersect $y = (x + 3)^{2}$

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Answer:

Correct Answer: 6. (a, d)

Solution:

Given,

$(x^{2} + x y + 4 x + 2 y + 4) \frac{d y}{d x} - y^{2} = 0$

$\Rightarrow [(x^{2} + 4 x + 4) + y (x + 2)] \frac{d y}{d x} - y^{2} = 0$

$\Rightarrow [(x + 2)^{2} + y (x + 2)] \frac{d y}{d x} - y^{2} = 0$

Put $x + 2 = X$ and $y = Y$ , then

$(X^{2} + X Y) \frac{d Y}{d X} - Y^{2} = 0$

$\Rightarrow X^{2} d Y + X Y d Y - Y^{2} d X = 0$

$\Rightarrow X^{2} d Y + Y (X d Y - Y d X) = 0$

$\Rightarrow - \frac{d Y}{Y} = \frac{X d Y - Y d X}{X^{2}}$

$\Rightarrow - d (\log | Y |) = d \frac{Y}{X}$

On integrating both sides, we get

$- \log | Y | = \frac{Y}{X} + C, where x + 2 = X and y = Y$

$\Rightarrow - \log | y | = \frac{y}{x + 2} + C$ …….(i)

Since, it passes through the point $(1, 3)$ .

$∴ - \log 3 = 1 + C$

$\Rightarrow C = - 1 - \log 3 = - (\log e + \log 3)$

$= - \log 3 e$

$∴$ Eq. (i) becomes

$\log | y | + \frac{y}{x + 2} - \log (3 e) = 0$

$\Rightarrow \log \frac{| y |}{3 e} + \frac{y}{x + 2} = 0$ …….(ii)

Now, to check option (a), $y = x + 2$ intersects the curve.

$\Rightarrow \log \frac{| x + 2 |}{3 e} + \frac{x + 2}{x + 2} = 0$

$\Rightarrow \log \frac{| x + 2 |}{3 e} = - 1$

$\Rightarrow \frac{| x + 2 |}{3 e} = e^{- 1} = \frac{1}{e}$

$\Rightarrow | x + 2 | = 3 or x + 2 = \pm 3$

$∴ x = 1, - 5$ (rejected), as $x > 0$

[given]

$∴ x = 1$ only one solution.

Thus, (a) is the correct answer.

To check option (c), we have

$y = (x + 2)^{2} and \log \frac{| y |}{3 e} + \frac{y}{x + 2} = 0$

$\Rightarrow \log \frac{| x + 2 |^{2}}{3 e} + \frac{(x + 2)^{2}}{x + 2} = 0 \Rightarrow \log \frac{| x + 2 |^{2}}{3 e} = - (x + 2)$

$\Rightarrow \frac{(x + 2)^{2}}{3 e} = e^{- (x + 2)} or (x + 2)^{2} \cdot e^{x + 2} = 3 e \Rightarrow e^{x + 2} = \frac{3 e}{(x + 2)^{2}}$

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Clearly, they have no solution.

To check option (d), $y = (x + 3)^{2}$

i.e. $\log \frac{| x + 3 |^{2}}{3 e} + \frac{(x + 3)^{2}}{(x + 2)} = 0$

To check the number of solutions.

Let $g (x) = 2 \log (x + 3) + \frac{(x + 3)^{2}}{(x + 2)} - \log (3 e)$

$∴ g^{'} (x) = \frac{2}{x + 3} + \frac{(x + 2) \cdot 2 (x + 3) - (x + 3)^{2} \cdot 1}{(x + 2)^{2}} - 0$

$= \frac{2}{x + 3} + \frac{(x + 3) (x + 1)}{(x + 2)^{2}}$

Clearly, when $x > 0$ , then, $g^{'} (x) > 0$

$∴ g (x)$ is increasing, when $x > 0$ .

Thus, when $x > 0$ , then $g (x) > g (0)$

$g (x) > \log \frac{3}{e} + \frac{9}{4} > 0$

Hence, there is no solution. Thus, option (d) is true.

Differential Equations 3 Question 6

6.

Answer:

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Differential Equations 3 Question 6

6.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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