#int dy/(4y) = int dx/(1+x)# ? : y = ?

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Answer 1 · 2017-05-23T18:12:48

#y = C_3 (x+1)^4#

Integrating both sides

#1/4log y = log(x+1)+C_1# or

#log y = 4log(x+1)+C_2# or

#y = C_3 (x+1)^4#

Answer 2 · 2017-05-23T18:19:24

#y = k(x+1)^4; k > 0#

Given: #int dy/(4y) = int dx/(1+x)#

Multiply both sides by 4:

#int dy/(y) = 4int dx/(1+x)#

Integrate:

#ln|y|=4ln|x+1|+C#

Move the 4 inside as a power:

#ln|y|= ln((x+1)^4)+C#

Make both sides the exponent of e, because this will eliminate the natural logarithms:

#e^(ln|y|) = e^(ln((x+1)^4)+C) = (e^C)(e^ln((x+1)^4))#

#e^C# is just an arbitrary constant #k# that must be greater than 0:

#y = k(x+1)^4; k > 0#