Solve the differential equation $dy/dx = 5x sqrt(y)$ where y!=0?

Question

$2sqrty = (5x^2)/2 +C$
$y= (25x^4+20Cx^2+4C^2)/16$

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Answer 1 · 2017-03-06T14:10:52

$y= (25 x^4)/16 + (5Cx^2)/4 + C^2/4$

This is "separable":

$dy/dx = 5x sqrty$

$1/ sqrty dy/dx = 5x$

Integrate both sides wrt x:

$int 1/ sqrty dy/dx dx =int 5x dx$

$int 1/ sqrty dy =5int x dx$

Integrate:

$2 sqrty = (5x^2)/2 + C$

$sqrty = 1/2((5x^2)/2 + C)$

$y = 1/4( (25 x^4)/4 + 5C x^2 + C^2)$

$= (25 x^4)/16 + (5Cx^2)/4 + C^2/4$