SATHEE: Differential Equations 2 Question 9

Differential Equations 2 Question 9

9. Let $y = y (x)$ be the solution of the differential equation, $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) = \log_{e} 4 - 1$ , then $y (e)$ is equal to

(2019 Main, 12 Jan I)

(a) $- \frac{e}{2}$

(b) $- \frac{e^{2}}{2}$

(d) $\frac{e^{2}}{4}$

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Answer:

Correct Answer: 9. (c)

Solution:

Given differential equation is

$\begin{aligned} x \frac{d y}{d x} + y = x \log_{e} x, (x > 1) \\ \Rightarrow & \frac{d y}{d x} + \frac{1}{x} y & = \log_{e} x \end{aligned}$

Which is a linear differential equation.

So, if $= e^{\int \frac{1}{x} d x} = e^{\log_{e} x} = x$

Now, solution of differential Eq. (i), is

$\begin{array}{cc} y \times x = \int (\log_{e} x) x d x + C \\ \Rightarrow & y x = \frac{x^{2}}{2} \log_{e} x - \int \frac{x^{2}}{2} \times \frac{1}{x} d x + C \\ \Rightarrow & [using integrat \\ y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} + C \end{array}$

$[using integration by parts]$

Given that, $2 y (2) = \log_{e} 4 - 1$

On substituting, $x = 2$ , in Eq. (ii),

we get

$2 y (2) = \frac{4}{2} \log_{e} 2 - \frac{4}{4} + C,$

[where, $y (2)$ represents value of $y$ at $x = 2$ ]

$\Rightarrow 2 y (2) = \log_{e} 4 - 1 + C$

$[∵ m \log a = \log a^{m}]$ From Eqs. (iii) and (iv), we get

$C = 0$

So, required solution is

$\begin{aligned} y x & = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} \\ x & = e, e y (e) = \frac{e^{2}}{2} \log_{e} e - \frac{e^{2}}{4} \end{aligned}$

Now, at

[where, $y (e)$ represents value of $y$ at $x = e$ ]

$\Rightarrow y (e) = \frac{e}{4} [∵ \log_{e} e = 1]$

Differential Equations 2 Question 9

9. Let $y = y (x)$ be the solution of the differential equation, $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) = \log_{e} 4 - 1$ , then $y (e)$ is equal to

Answer:

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Differential Equations 2 Question 9

9. Let y=y(x) be the solution of the differential equation, xdydx+y=xloge⁡x,(x>1). If 2y(2)=loge⁡4−1, then y(e) is equal to

Answer:

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9. Let $y = y (x)$ be the solution of the differential equation, $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) = \log_{e} 4 - 1$ , then $y (e)$ is equal to