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

##### 1 Answer

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

#### Explanation:

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,

# 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,

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

Thus the solution is:

# \ x^2y = x^4 -1#

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