If #dy/dx# =#3x#+#2/x^2# and y=2 when #x#=1, find the function y?

1 Answer
Noah G
Feb 6, 2018

#y = 3/2x^2 - 2/x + 5/2#

Explanation:

The nice thing about this differential equation is that the #dy/dx# is already isolated, therefore the answer can be obtained by simply integrating both sides.

#int dy = int 3x + 2/x^2 dx#

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

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

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

#4 - 3/2 = C#

#5/2 = C#

Therefore, the solution is #y = 3/2x^2 - 2/x + 5/2#

Hopefully this helps!

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