Problem 02 | Linear Equations

Problem 02
$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$
 

Solution 02
$2(2xy + 4y - 3) \, dx + (x + 2)^2 \, dy = 0$

$2[ \, 2y(x + 2) - 3 \, ] \, dx + (x + 2)^2 \, dy = 0$

$\dfrac{2[ \, 2y(x + 2) - 3 \, ] \, dx}{(x + 2)^2 \, dx} + \dfrac{(x + 2)^2 \, dy}{(x + 2)^2 \, dx} = 0$

$\dfrac{4y(x + 2) - 6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

$\dfrac{4y(x + 2)}{(x + 2)^2} - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

$\left( \dfrac{4}{x + 2} \right) y - \dfrac{6}{(x + 2)^2} + \dfrac{dy}{dx} = 0$

Problem 01 | Linear Equations

Problem 01
$(x^5 + 3y) \, dx - x \, dy = 0$
 

Solution 01
$(x^5 + 3y) \, dx - x \, dy = 0$

$\dfrac{(x^5 + 3y) \, dx}{x \, dx} - \dfrac{x \, dy}{x \, dx} = 0$

$\dfrac{x^5 + 3y}{x} - \dfrac{dy}{dx} = 0$

$x^4 + \dfrac{3y}{x} - \dfrac{dy}{dx} = 0$

$\dfrac{dy}{dx} - \dfrac{3}{x}y = x^4$       → linear in y
 

Hence,
$P = -\dfrac{3}{x}$

$Q = x^4$
 

Integrating factor:
$e^{\int P \, dx} = e^{\int \left( -\frac{3}{x} \right)dx}$

$e^{\int P \, dx} = e^{-3\int \frac{dx}{x}}$

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