# differential Equation

2 posts / 0 new
Wilfredo Milla
differential Equation

substitution suggested by equation
1. (3siny - 5x)dx + 2x^2cotydy = 0
2. (ke^2v - u)du = 2e^2v(e^2v + ku)dv = 0

Jhun Vert

Number 1
$(3\sin y - 5x) \, dx + 2x^2 \cot y \, dy = 0$

$\left( \dfrac{3}{\csc y} - 5x \right) dx + 2x^2 \cot y \, dy = 0$

$(3 - 5x \csc y) \, dx + 2x^2 \csc y \cot y \, dy = 0$

$(3 - 5x \csc y) \, dx - 2x^2 (-\csc y \cot y \, dy) = 0$

Let
$z = \csc y$

$dz = -\csc y \cot y \, dy$

Thus,
$(3 - 5xz) \, dx - 2x^2dz = 0$

$3 - 5xz - 2x^2\dfrac{dz}{dx} = 0$

$\dfrac{3}{2x^2} - \dfrac{5x}{2x^2}z - \dfrac{dz}{dx} = 0$

$\dfrac{dz}{dx} + \dfrac{5}{2x}z = \dfrac{3}{2x^2}$   ←   linear equation of order one

Integrating factor:
$e^{\int \frac{5}{2x} \, dx} = e^{\frac{5}{2} \ln x} = e^{\ln x^{5/2}} = x^{5/2}$

Hence
$\displaystyle z(x^{5/2}) = \int \left( \dfrac{3}{2x^2} \right)(x^{5/2}) \, dx + C$

$\displaystyle z(x^{5/2}) = \dfrac{3}{2}\int x^{1/2} \, dx + C$

$z(x^{5/2}) = \dfrac{3}{2} \left( \dfrac{x^{3/2}}{3/2} \right) + C$

$x^{5/2} \csc y = x^{3/2} + C$       answer

You may recheck my solution, I did not run any check on it.

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