First slide

Methods of solving first order first degree differential equations

Question

$Let y (x) be the solution of the differential equation (xlogx) \frac{dy}{dx} + y = 2 xlogx, (x \geq 1) then y (e) is$

Moderate

A

e
B

0
C

2
D

2e

Solution

$\begin{array}{l} \frac{d y}{d x} + \frac{1}{x \ln x} y = 2 \\ I.F. = e^{\int \frac{1}{x \ln x} d x} = e^{\ln (\ln x)} = \ln x \\ Sol of the D.E is y (\ln x) = \int 2 \ln x d x = 2 [\ln x \int 1 d x - \int (\frac{1}{x} \int 1 d x) d x] \\ y (\ln x) = 2 (x \ln x - \int 1 d x) \\ y (\ln x) = 2 x (\ln x - 1) + c put x = 1 \\ 0 = - 2 + c \to c = 2 \\ y (\ln x) = 2 x (\ln x - 1) + 2 \\ When x = e \\ y = 2 e (1 - 1) + 2 \end{array}$
$y = 2$

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