How do you solve #xy'+2y=4x^2# given y(1)=0?

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Answer 1 · 2017-05-09T00:48:55

# y = x^2 -1/x^2#

We have:

# xy'+2y=4x^2 #

Which we can write as:

# y'+2/xy=4x \ \ \ \ ...... [1]#

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) #

Then the integrating factor is given by;

# IF = e^(int P(x) dx) #
# " " = exp(int \ 2/x \ dx) #
# " " = exp( 2lnx ) #
# " " = e^( lnx^2 ) #
# " " = x^2 #

And if we multiply the DE [1] by this Integrating Factor, #IF#, we will have a perfect product differential;

# y'+2/xy=4x #

# :. x^2y'+2xy=4x^3 #

# :. d/dx (x^2y) = 4x^3 #

Which we can directly integrate to get:

# int \ d/dx (x^2y) \ dx = int \ 4x^3 \ dx#
# :. x^2y = x^4 + C#

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

# 1*0 = 1 + C => C = -1#

Thus the solution is:

# \ x^2y = x^4 -1#
# :. y = x^2 -1/x^2#