Solving a first order linear differential equation by variation of parameters

[bitrex]

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.

 

[djeitnstine]

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]

Double post

 

[rock.freak667]

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.

 

[bitrex]

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]

-2-1= -3, not -1.


The derivative of x^-2 is -2x^-3, not -2x^-1.

 

[bitrex]

:facepalm: Yeah, that's the problem. Getting my differentiation and integration confused again.