substitution suggested by equation 1. (3siny - 5x)dx + 2x^2cotydy = 0 2. (ke^2v - u)du = 2e^2v(e^2v + ku)dv = 0
Number 1 $(3\sin y - 5x) \, dx + 2x^2 \cot y \, dy = 0$
$\left( \dfrac{3}{\csc y} - 5x \right) dx + 2x^2 \cot y \, dy = 0$
$(3 - 5x \csc y) \, dx + 2x^2 \csc y \cot y \, dy = 0$
$(3 - 5x \csc y) \, dx - 2x^2 (-\csc y \cot y \, dy) = 0$
$dz = -\csc y \cot y \, dy$
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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Number 1
$(3\sin y - 5x) \, dx + 2x^2 \cot y \, dy = 0$
$\left( \dfrac{3}{\csc y} - 5x \right) dx + 2x^2 \cot y \, dy = 0$
$(3 - 5x \csc y) \, dx + 2x^2 \csc y \cot y \, dy = 0$
$(3 - 5x \csc y) \, dx - 2x^2 (-\csc y \cot y \, dy) = 0$
$z = \csc y$
$dz = -\csc y \cot y \, dy$
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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