# What is the solution of the Differential Equation # x(dy/dx)+3y+2x^2=x^3+4x #?

##### 1 Answer

# y = (5x^6 - 12x^5+30x^4 + C)/(30x^3) #

#### Explanation:

We have:

# x(dy/dx)+3y+2x^2=x^3+4x #

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 rewrite the equations in standard form as:

# (dy/dx)+3/xy = x^2-2x+4 ..... [1] #

Then the integrating factor is given by;

# I = e^(int P(x) dx) #

# \ \ = exp(int \ 3/x \ dx) #

# \ \ = exp( 3lnx ) #

# \ \ = exp( lnx^3 ) #

# \ \ = x^3 #

And if we multiply the DE [1] by this Integrating Factor,

# (dy/dx)x^3+3x^2y = x^5-2x^4+4x^3 #

# :. d/dx (x^3y) = x^5-2x^4+4x^3 #

Which we can directly integrate to get:

# x^3y = int \ x^5-2x^4+4x^3 \ dx #

# :. x^3y = x^6/6-(2x^5)/5+x^4 + c #

# :. x^3y = (5x^6 - 12x^5+30x^4 + 30c)/30 #

# :. y = (5x^6 - 12x^5+30x^4 + C)/(30x^3) #

Confirming the given answer.