My question is: "find the particular solution y(x) of the given differential equation which complies with the initial condition" # xy'+x^2=2y# with #y(1)=0# How to solve it? Thank you!

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Answer 1 · 2018-05-18T14:59:20

# y = -x^2 \ ln |x| #

We have:

# xy'+x^2=2y# with #y(1)=0#

We can use an integrating factor when we have a First Order Linear non-homogeneous Ordinary Differential Equation of the form;

# dy/dx + P(x)y=Q(x) #

So, we can put the equation in standard form:

# xy'-2y = -x^2#

# :. y'- (2y)/x = -x# ..... [A]

Then the integrating factor is given by;

# I = e^(int P(x) dx) #
# \ \ = exp(int \ -2/x \ dx) #
# \ \ = exp(-2ln|x| ) #
# \ \ = e^(ln|x|^(-2)) #
# \ \ = 1/x^2 #

And if we multiply the DE [A] by this Integrating Factor, #I#, we will have a perfect product differential (in fact the original equation);

# :. 1/x^2 \ y' - 1/x^2 \ (2y)/x = 1/x^2 \ (-x) #

# :. 1/x^2 \ y' - 2/x^3 \ y = -1/x #

# :. d/dx (y/x^2) = -1/x #

This is now separable, so by "separating the variables" we get:

# y/x^2 = int -1/x \ dx + C #

Which is trivial to integrate to get the General Solution:

# y/x^2 = -ln | x| + C #

Applying the initial condition #y(1)=0# we get:

# 0/1 = -ln | 1| + C => C = 0#

Leading to the Particular Solution:

# y = x^2(-ln | x| + 0) #

# :. y = -x^2 \ ln |x| \ \ \ # QED