Question

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

Answer 1

`y =cx+x^2`

Explanation:

Solve first the homogeneous equation that is separable:

`dy/dx = y/x`

`dy/y = dx/x`

Integrate both sides:

`int dy/y = int dx/x`

`ln abs y = ln abs x +C`

and taking the exponential of both sides:

`y = cx`

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

`bar y = c(x) x`

Differentiate using the product rule:

`(dbary)/dx = x c'(x) + c(x)`

and substitute in the original equation:

`(dbary)/dx =(bary)/x +x`

`x c'(x) + c(x) = (c(x) x)/x +x`

`x c'(x) + c(x) = c(x)+x`

`x c'(x) = x`

`c'(x) = 1`

`int c'(x)dx = int dx`

`c(x) = x`

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

`bar y = c(x) x = x^2`

Then the complete solution of the equation is:

`y =cx+x^2`

In fact:

`dy/dx = c+2x = (c+x)+x = (cx+x^2)/x+x =y/x +x`

Answer 2

`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, `I`, we will have a perfect product differential;

`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`