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#