# What is a solution to the linear differential equation: #dy/dx= y/x+x#?

##### 2 Answers

#### Explanation:

Solve first the homogeneous equation that is separable:

Integrate both sides:

and taking the exponential of both sides:

Use now the variable coefficient method to find a solution to the compete equation in the form;

Differentiate using the product rule:

and substitute in the original equation:

and we do not need the constant because we can choose just one solution:

Then the complete solution of the equation is:

In fact:

# y = x^2 +Cx #

#### Explanation:

We have:

# dy/dx = y/x+x #

Which we can write as:

# dy/dx +(-1/x) y = x #

Which is of the form:

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

So, we can construct an Integrating Factor:

# I = exp(int \ P(x) \ dx #

# \ \ = exp(int \ -1/x \ dx) #

# \ \ = exp(-lnx) #

# \ \ = 1/x #

And if we multiply the DE by this Integrating Factor,

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

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

This is now separable, so we can "separate the variables" to get:

# y/x = int \ 1 \ dx #

Which is trivial to integrate:

# y/x = x+C #

Leading to the GS:

# y = x^2 +Cx #