MHB Problem with linear differential equation

[stripedcat]

Still learning the formatting commands, sorry!

I'm aware of the $(dy/dx) + P(x)y=Q(x)$ formula, as well as the $e^{\int P(x) dx}$ formula needed to get the "I" factor.

Here's the equation.

$$(dy/dx)+(2/x)y=3x-5$$

The "$P(x)$" would be $(2/x), \int 2/x\ dx = 2 \ln(x), e^{2 \ln(x)} = x^2$, so $x^2$ is the "I factor", and I know this gets multiplied by both sides
$$x^2((dy/dx)+(2/x)y=x^2(3x-5)$$

$3x^3+5x^2$ simplification on the right.

I'm not sure what to do with the left side. I know I need to get that dx out of there so it can go over to the right.

I think I need to start isolating the dy/dx expression...
$$x^2((dy/dx) = (3x^3+5x^2)-((2/x)y)$$
$$dy/dx = ((3x^3+5x^2)-((2/x)y))/x^2$$

This doesn't seem right at all.

 

[topsquark]

Still learning the formatting commands, sorry!

I'm aware of the $(dy/dx) + P(x)y=Q(x)$ formula, as well as the $e^{\int P(x) dx}$ formula needed to get the "I" factor.

Here's the equation.

$$(dy/dx)+(2/x)y=3x-5$$

The "$P(x)$" would be $(2/x), \int 2/x\ dx = 2 \ln(x), e^{2 \ln(x)} = x^2$, so $x^2$ is the "I factor", and I know this gets multiplied by both sides
$$x^2((dy/dx)+(2/x)y=x^2(3x-5)$$

$3x^3+5x^2$ simplification on the right.

I'm not sure what to do with the left side. I know I need to get that dx out of there so it can go over to the right.

I think I need to start isolating the dy/dx expression...
$$x^2((dy/dx) = (3x^3+5x^2)-((2/x)y)$$
$$dy/dx = ((3x^3+5x^2)-((2/x)y))/x^2$$

This doesn't seem right at all.

The x^2 is an integrating factor. So the left hand side is now in a form that you can write:
[math]\frac{d}{dx} \left ( x^2 y \right ) = 2x y + x^2 \frac{dy}{dx}[/math]

Thus your original equation is now
[math]\frac{d}{dx} \left ( x^2 y \right ) = x^2 (3x - 5)[/math]

How do you finish this?

-Dan

 

[stripedcat]

The x^2 is an integrating factor. So the left hand side is now in a form that you can write:
[math]\frac{d}{dx} \left ( x^2 y \right ) = 2x y + x^2 \frac{dy}{dx}[/math]

Thus your original equation is now
[math]\frac{d}{dx} \left ( x^2 y \right ) = x^2 (3x - 5)[/math]

How do you finish this?

-Dan

Yes, I was just looking at this. I didn't know about making it into d/dx and yanking the y out to multiply by the I factor. Give me a minute and I will edit this post with what I come up with.

EDIT: And here's what I came up with

Integrate both sides will cancel the d/dx.

$$x^2y=\int 3x^3-5x^2$$

I don't know the command for 'something over something', I was guessing on the integral symbol

$$x^2y=(3x^4/4)-(5x^3/3)+C$$

How's that?

EDIT: Yes I know, it keeps going after that in order to get just the y

 

[Prove It]

\frac{}{} gives the fraction template. You type whatever you want in the numerator in the first set of curly brackets, and whatever you want in the denominator in the second set of curly brackets.

 

[topsquark]

Yes, I was just looking at this. I didn't know about making it into d/dx and yanking the y out to multiply by the I factor. Give me a minute and I will edit this post with what I come up with.

EDIT: And here's what I came up with

Integrate both sides will cancel the d/dx.

$$x^2y=\int 3x^3-5x^2$$

I don't know the command for 'something over something', I was guessing on the integral symbol

$$x^2y=(3x^4/4)-(5x^3/3)+C$$

How's that?

EDIT: Yes I know, it keeps going after that in order to get just the y

Looks good to me. :)

-Dan