How do I solve this differential equation?

Question

#y'=y^2/(1+sqrtx)#

for #x>=0#

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Answer 1 · 2018-03-23T19:04:28

Use the separation of variables method.

#y'=y^2/(1+sqrtx)#

Write #y'# as #dy/dx#

#dy/dx=y^2/(1+sqrtx)#

Separate variables by multiplying both sides by #dx/y^2#:

#dy/y^2 = 1/(1+sqrtx)dx#

Integrate both sides:

#-1/y = 2sqrtx-2ln(sqrtx+1)+C#

#y = 1/(2ln(sqrtx+1)-2sqrtx+C)#