How do you solve it ? (x−4)y′=5y+3

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Answer 1 · 2018-04-06T20:24:26

Write #y'# as #dy/dx#:

#(x-4)dy/dx = 5y+3#

Use the separation of variables method:

#dy/(5y+3)=dx/(x-4)#

Integrate both sides:

#int dy/(5y+3)= int dx/(x-4)#

#1/5ln(5y+3) = ln(x-4) +C #

#ln(5y+3) = 5ln(x-4)+ C#

#ln(5y+3) = ln((x-4)^5)+ C#

#e^(ln(5y+3)) = e^(ln((x-4)^5)+ C)#

#e^ln(5y+3) = e^Ce^(ln((x-4)^5))#

#5y+3 = C_1(x-4)^5#

#5y = C_1(x-4)^5-3#

#y = C_2(x-4)^5-3/5#

How do you solve it ? (x−4)*y′=5*y+3