# What is the general solution of the differential equation? : # dy/dx + 3y = 3x^2e^(-3x) #

##### 2 Answers

#### Explanation:

Note: This differential equation is first order and linear one.

1) I multiplied both sides with

2) I integrated both sides.

3) I multiplied with

# y = x^3e^(-3x) + Ce^(-3x) #

#### Explanation:

We have:

# dy/dx + 3y = 3x^2e^(-3x) # ..... [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;

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

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

# \ \ = exp( 3x ) #

# \ \ = e^(3x) #

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

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

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

Which we can now directly integrate to get:

# e^(3x)y = int \ x^3 \ dx #

# :. e^(3x)y = x^3+ C #

# :. y = (x^3+ C)/e^(3x) #

# :. y = x^3e^(-3x) + Ce^(-3x) #