SATHEE: Differential Equations 2 Question 2

Differential Equations 2 Question 2

2. Consider the differential equation, $y^{2} d x + x - \frac{1}{y} d y = 0$ . If value of $y$ is 1 when $x = 1$ , then the value of $x$ for which $y = 2$ , is

(2019 Main, 12 April I)

(a) $\frac{5}{2} + \frac{1}{\sqrt{e}}$

(b) $\frac{3}{2} - \frac{1}{\sqrt{e}}$

(d) $\frac{3}{2} - \sqrt{e}$

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Answer:

Correct Answer: 2. (b)

Solution:

Given differential equation is

$y^{2} d x + x - \frac{1}{y} d y = 0$

$\Rightarrow \frac{d x}{d y} + \frac{1}{y^{2}} x = \frac{1}{y^{3}}$ , which is the linear differential equation of the form $\frac{d x}{d y} + P x = Q$ .

Here, $P = \frac{1}{y^{2}}$ and $Q = \frac{1}{y^{3}}$

Now, IF $= e^{\int \frac{1}{y^{2}} d y} = e^{- \frac{1}{y}}$

$∴$ The solution of linear differential equation is

$\begin{aligned} x \cdot (I F) & \int (I F) d y + C \\ \Rightarrow x e^{- 1 / y} & = \int \frac{1}{y^{3}} e^{- 1 / y} d y + C \\ ∴ x e^{- 1 / y} & = \int (- t) e^{t} d t + C [∵ let - \frac{1}{y} = t \Rightarrow + \frac{1}{y^{2}} d y = d t] \\ = - t e^{t} + \int e^{t} d t + C [integration by parts] \\ = - t e^{t} + e^{t} + C \end{aligned}$

$\Rightarrow x e^{- 1 / y} = \frac{1}{y} e^{- 1 / y} + e^{- 1 / y} + C$

Now, at $y = 1$ , the value of $x = 1$ , so

$1 \cdot e^{- 1} = e^{- 1} + e^{- 1} + C \Rightarrow C = - \frac{1}{e}$

On putting the value of $C$ , in Eq. (i), we get

$x = \frac{1}{y} + 1 - \frac{e^{1 / y}}{e}$

So, at $y = 2$ , the value of $x = \frac{1}{2} + 1 - \frac{e^{1 / 2}}{e} = \frac{3}{2} - \frac{1}{\sqrt{e}}$

Differential Equations 2 Question 2

2. Consider the differential equation, $y^{2} d x + x - \frac{1}{y} d y = 0$ . If value of $y$ is 1 when $x = 1$ , then the value of $x$ for which $y = 2$ , is

Answer:

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Differential Equations 2 Question 2

2. Consider the differential equation, y2dx+x−1ydy=0. If value of y is 1 when x=1, then the value of x for which y=2, is

Answer:

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2. Consider the differential equation, $y^{2} d x + x - \frac{1}{y} d y = 0$ . If value of $y$ is 1 when $x = 1$ , then the value of $x$ for which $y = 2$ , is