Question

solve differential equation y- x * dy/dx=3(1+x^2 dy/dx)​

Answer 1

EXPLANATION.

$\sf \implies \bigg( y - x \dfrac{dy}{dx} \bigg) = 3 \bigg( 1 + x^{2} \dfrac{dy}{dx} \bigg).$

As we know that,

$\sf \implies \bigg( y - x \dfrac{dy}{dx} \bigg) = \bigg(3 + 3 x^{2} \dfrac{dy}{dx} \bigg)$

$\sf \implies \bigg(3x^{2} + x \bigg) \dfrac{dy}{dx} = (y - 3).$

$\sf \implies \dfrac{dy}{(y - 3)} = \dfrac{dx}{(3x^{2} + x)}$

Integrate both sides, we get.

$\sf \implies \displaystyle \int \dfrac{dy}{(y - 3)} = \int \dfrac{dx}{(3x^{2} + x)}$

$\sf \implies \displaystyle \int \bigg( \dfrac{dx}{(3x^{2} + x)} \bigg)$

$\sf \implies \displaystyle \int \dfrac{dx}{x^{2} \bigg( 3 + \dfrac{1}{x} \bigg)}$

By using substitution method, we get.

$\sf \implies \displaystyle 3 + \dfrac{1}{x} = t.$

Differentiate w.r.t x, we get.

$\sf \implies \displaystyle-\dfrac{1}{x^{2} } dx = dt.$

$\sf \implies \displaystyle \int \dfrac{-dt}{t} = - ln|t| = - ln \bigg| 3 + \frac{1}{x} \bigg|$

$\sf \implies \displaystyle \int \dfrac{dy}{(y - 3)} = \int \dfrac{dx}{(3x^{2} + x)}$

$\sf \implies \displaystyle ln(y - 3) = -ln \bigg( 3 + \dfrac{1}{x} \bigg) + ln(k).$

$\sf \implies \displaystyle ln(y - 3) = -ln \bigg( \dfrac{3x + 1}{x} \bigg) + ln(k).$

                                                                                                                   

MORE INFORMATION.

Some important results (solution by inspection).

$\sf \displaystyle (1) = d(x y) = x dy + y dx.$

$\sf \displaystyle (2) = d \bigg( \dfrac{x}{y} \bigg) = \dfrac{y dx - x dy}{y^{2} } , \ y \ne 0.$

$\sf \displaystyle (3) = d \bigg(\dfrac{y}{x}\bigg) = \dfrac{x dy - y dx }{x^{2} } , \ x \ne 0.$

$\sf \displaystyle (4) = d \bigg( \dfrac{x^{2} }{y} \bigg) = \dfrac{2xydx - x^{2} dy}{y^{2} } , \ y \ne 0.$

$\sf \displaystyle (5) = d \bigg(\dfrac{y^{2} }{x} \bigg) = \dfrac{2xydy - y^{2} dy}{x^{2} }$

$\sf \displaystyle (6) = d \bigg(\dfrac{x^{2} }{y^{2} } \bigg) = \dfrac{2xy^{2} dx - 2x^{2} ydy}{y^{4} }$

$\sf \displaystyle (7) = d \bigg( tan^{-1} \bigg(\dfrac{y}{x} \bigg)\bigg) = \dfrac{xdy - ydx}{x^{2} + y^{2} }$

$\sf \displaystyle (8) = d \bigg( log_{e} (xy) \bigg) = \dfrac{xdy + ydx}{xy}$

$\sf \displaystyle (9) = d \bigg( log_{e} \bigg( \dfrac{x}{y} \bigg) \bigg) = \dfrac{ydx - xdy}{xy}$