Question

(x-2)y'+y=(x^2-4)
linear equation differential equation​

Answer 1

Step-by-step explanation:

We have,

$(x - 2) \frac{dy}{dx} + y = ( {x}^{2} - 4)$

Dividing both side by (x - 2)

$\implies \: \frac{dy}{dx} + \frac{y}{ (x - 2 )} = (x + 2)$

$I.F. = {e}^{ \int \frac{dx}{x - 2} } = {e}^{ ln(x - 2) } = (x - 2)$

$y(x - 2) = \int(x + 2)(x - 2)dx$

$\implies \: y(x - 2) = \int(x^{2} - 4)dx$

$\implies \: y(x - 2) = \frac{ {x}^{3} }{3} - 4 x + c$