$dy/dx = sqrt(x/y)$ ?

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Answer 1 · 2018-04-26T22:08:20

$y=pmsqrt(4/3x^(3//2)+C)$

We have the differential equation:

$dy/dx=sqrtx/y$

Use separation of variables to get $x$ and $y$ on opposite sides of the equation:

$ycolor(white).dy=sqrtxcolor(white).dx$

Now integrate both sides:

$intycolor(white).dy=intx^(1//2)color(white).dx$

$1/2y^2=1/(3//2)x^(3//2)+C$

$1/2y^2=2/3x^(3//2)+C$

$y^2=4/3x^(3//2)+C$

(I'll just keep writing $C$ to represent any arbitrary constant, it doesn't matter that we've multiplied it by two.)

$y=pmsqrt(4/3x^(3//2)+C)$

Answer 2 · 2018-04-27T00:58:25

$y^(3/2) = x^(3/2) + C$

We have:

$dy/dx = sqrt(x/y)$

We can collect terms:

$dy/dx = sqrt(x)/sqrt(y)$

$sqrt(y) \ dy/dx = sqrt(x)$

Which is a Separable DE, so we can "separate the variables" to get

$int \ sqrt(y) \ dy = int \ sqrt(x) \ dx$

Now we can integrate:

$(y^(3/2))/(3/2) = (x^(3/2))/(3/2) + c$

Leading to the General Solution:

$y^(3/2) = x^(3/2) + C$

dy/dx = sqrt(x/y)dy/dx = sqrt(x/y) ?