Question

solve differential equation
y-xdy/dx=a(y^2+dydx)
​

Answer 1

EXPLANATION.

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

As we know that,

We can write equation as,

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

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

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

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

Integrate both sides of the equation, we get.

$\implies \displaystyle \int \dfrac{dy}{y(1 - ay)} \ = \int \dfrac{dx}{x + a}$

$\implies \displaystyle \int \dfrac{(1 - ay) + ay }{y(1 - ay)} dy \ = \int \dfrac{dx}{x + a}$

$\implies \displaystyle \int \dfrac{(1 - ay)}{y(1 - ay)} dy \ + \int \dfrac{ay}{y(1 - ay)} dy \ = \int \dfrac{dx}{a + x}$

$\implies \displaystyle \int \dfrac{dy}{y} \ + \int \dfrac{a}{(1 - ay)} dy \ = \int \dfrac{dx}{x + a}$

$\implies \displaystyle ln|y| - ln|1 - ay| = ln|x + a| + ln|C|$

$\implies \displaystyle \dfrac{y}{(1 - ay)} \ = C(x + a)$

                                                                                                                       

MORE INFORMATION.

Differential equations of the form : dy/dx = f(x).

⇒ ∫dy = ∫f(x)dx + c.

Differential equations of the form dy/dx = f(x) g(y).

⇒ ∫dy/g(y) = ∫f(x)dx + c.