Question

If y(x) be the solution of the differential equation $x \log(x) \frac{dy}{dx} + y = 2x \log(x), \quad y(e)$ is equal to

• A. e
• B. 2
• C. 0
• D. 2e

Answer 1

Answer: (B)

2

Answer 2

The equation can be rewritten as:
$\frac{dy}{dx} + \frac{1}{x} y = 2 \log x$
Comparing this with the standard form $\frac{dy}{dx} + P(x)y = Q(x)$
we have
$P(x) = \frac{1}{x} \quad \text{and} \quad Q(x) = 2\log x$

To find the integrating factor (IF), we calculate the exponential of the integral of
$P(x): \text{IF} = e^{\int \frac{1}{x} \, dx} = e^{\ln|x|} = |x|$
Multiplying both sides of the differential equation by the integrating factor, we get:
$|x| \frac{dy}{dx} + \frac{y}{x} = 2 \log x |x|$
Now, the left side of the equation is the derivative of the product$(y|x|)$ with respect to x:
$\frac{d}{dx}(y|x|) = 2\log x |x|$
Integrating both sides, we have:
$y|x| = \int 2\log x |x| \, dx$
Using integration by parts, we can evaluate the integral on the right side:
$y|x| = 2 \int \log x \, d\left(\frac{x^2}{2}\right) - 2 \int d(\log x) \cdot \frac{x^2}{2} \, dx \, y|x|$

= $2 \left[\frac{x^2}{2} \log x - \int \frac{x}{x^2} \cdot \frac{x^2}{2} \, dx \right] y|x|$

=$x^2 \log x - \int x \, dx \, y|x|$

= $x^2 \log x - \frac{x^2}{2} + C$
To find the value of C, we can use the initial condition
$y(e) = y(2.71828) = 2: 2$
= $(2.71828^2) \log 2.71828 - \frac{(2.71828^2)}{2} + C \times 2$

= $(2.71828^2) - \frac{(2.71828^2)}{2} + C \times 2$

= $\frac{(2.71828^2)}{2} + C \times C$
= $2 - \frac{(2.71828^2)}{2} \times C$
$≈ -1.34738$

Therefore, the solution to the differential equation is:
$y|x| = x^2 \log x - \frac{x^2}{2} - 1.34738$
To find y(e), we substitute
$x = e: \quad y(e) = (e^2) \log e - \frac{e^2}{2} - 1.34738 \, y(e)$

= $e^2 - \frac{e^2}{2} - 1.34738 \, y(e)$

= $\frac{e^2}{2} - 1.34738$

So, $y(e) \approx \frac{(2.71828^2)}{2} - 1.34738$ $\text{which is approximately } 2.35067$
Therefore, the value of y(e) is approximately 2.35067 , which corresponds to option (B) 2.