Question

x
logx dy/dx+ y = 2logx
​

Answer 1

EXPLANATION.

⇒ x ㏒(x) dy/dx + y = 2㏒(x).

As we know that,

Divide equation by x㏒(x), we get.

$\implies \dfrac{dy}{dx} + \dfrac{y}{x log(x)} = \dfrac{2 log(x)}{x log(x)}$

$\implies \dfrac{dy}{dx} + \bigg(\dfrac{1}{x log(x)} \bigg) y = \dfrac{2}{x}$

Now, the equation is in the form of linear differential equations, we get.

$\implies \dfrac{dy}{dx} + P(x) y = Q(x)$

$\implies e^{\int p(x)dx} = Integrating \ \ factor = I.F$

$\implies y \times I.F = \displaystyle \int \bigg[ Q(x) \times I.F \bigg] dx$

$\implies y \times e^{\int p(x)dx} = \displaystyle \int \bigg[Q(x) \times e^{\int p(x)dx} \bigg] dx$

Using this formula in the equation, we get.

$\implies e^{\displaystyle \int \dfrac{1}{x log(x)}dx }$

$\implies e^{log(logx)} = log(x)$

$\implies y \times log(x) = \displaystyle \int \bigg( log(x) \times \dfrac{2}{x} \bigg) dx$

$\implies y \times log(x) = 2 \displaystyle \int \dfrac{log(x)}{x} dx$

$\implies \displaystyle \int \dfrac{log(x)}{x} dx$

By applying substitution method, we get.

⇒ ㏒(x) = t.

Differentiate w.r.t x, we get.

⇒ dx/x = dt.

Put the values in the equation, we get.

$\implies \int t dt = \dfrac{t^{2} }{2} + C.$

Put the value of t = ㏒(x) in the equation, we get.

(㏒(x))²/2 + C.

Put the values in the main equation, we get.

$\implies y \times log(x) = 2 \times \dfrac{log(x)^{2} }{2} + C.$

$\implies y \times log(x) = log(x)^{2} + C.$