How do you solve the differential equation #dy/dx=6y^2x#, where #y(1)=1/25# ?

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Answer 1 · 2018-04-26T19:09:11

#=> y = 1/(28-3x^2)#

We are given:

#dy/dx = 6y^2x#

Separate the variables:

#1/y^2 dy = 6x dx#

Integrate both sides:

#int 1/y^2 dy = int 6x dx#

#-1/y = 3x^2 + C#

#1/y = -3x^2+C#

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

where #C# is an arbitrary constant of integration.

Now solve for #y(1)# to find #C#:

#y(1) = 1/25 = -1/(3(1)^2+C)#

#-1/25 = 1/(3+C)#

#3+C = -25#

#C = -28#

Hence, the final solution is:

#y = -1/(3x^2-28)#

#=> y = 1/(28-3x^2)#