What is the general solution of the differential equation # xyy'=x^2+1 # ?

1 Answer
Steve M
Mar 23, 2017

# y^2 = x^2 + 2ln|x| + A #

Explanation:

The differential equation

# xyy'=x^2+1 #

is a First Order linear separable Differential Equation which can be solved simply by rearranging and collection term in #x# on the RHS and term in #y# on the LHS;

# ydy/dx = (x^2+1)/x #

And now we "separate the variables" to get;

# int \ y \ dy = int \ x+1/x \ dx#

Which is trivial to integrate to get:

# \ \ 1/2y^2 = 1/2x^2 + ln|x| + C #

# :. y^2 = x^2 + 2ln|x| + 2C #
# :. y^2 = x^2 + 2ln|x| + A #

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