# Solving a first order linear differential equation by variation of parameters

## Homework Statement

I have to solve the following differential equation by the "variation of parameters" method.

## Homework Equations

$$\frac{dy}{dx}x +2y = 3x$$

## The Attempt at a Solution

The associated homogeneous equation of the initial equation is:

$$\frac{dy}{dx} = -2x^{-1}y$$

So

$$\frac{1}{y}dy = -2x^{-1}dx$$

$$ln(y) = -2ln(x)$$

$$ln(y) = ln(x^{-2})$$

$$y = x^{-2}$$

Unfortunately, this doesn't satisfy the homogeneous equation.

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djeitnstine
Gold Member
Where did the '3' go to? When you divide by x it is $$\frac{dy}{dx}=3-2\frac{y}{x}$$ not $$-2\frac{y}{x}$$

Then you do the substitution v=y/x, then it comes easy.

djeitnstine
Gold Member
Double post

rock.freak667
Homework Helper
You can use djeitnstine guide OR divide by x and then use an integrating factor. But you said you had a certain method to use.

Since that is the case then you should know that y1=x-2 is a solution of the homogenous equation. So just find the Wronskian of y1, which should be easy and just put it into the formula.

I'm sorry I didn't clarify in my earlier post - I'm supposed to take an equation of the form $$y' + p(x)y = q(x)$$, set the right hand side of my equation to zero, solve that equation by using separation of variables, and then take the homogeneous equation result as yh*u and substitute that back into the original differential equation, and get another separable equation. So I do the following with the answer to the homogeneous equation above and the original equation:

$$\frac{dy}{dx}(x^{-2}u) + 2x^{-1}x^{-2}u = 3$$

=

$$-2x^{-1}u + x^{-2}\frac{du}{dx} + 2x^{-3}u = 3$$

The way it's supposed to work is that the first term and the third term cancel, leaving me with a linear differential equation in du/dx to solve. However, in this case it doesn't seem to be working out that way...

HallsofIvy