# Solving an equation

1. May 29, 2005

### Icebreaker

$$y' + 2y = 6x + 5$$

Can I solve it by...

$$\frac{dy}{dx} + 2y = 6x + 5$$

$$dy = (6x + 5 - 2y)dx$$

$$y = 3x^2 + 5x - 2yx$$

$$y + 2yx = 3x^2 + 5x$$

$$y = \frac{3x^2 + 5x}{1 + 2x}$$

2. May 29, 2005

### arildno

No you can't.
1) Why do you think this works?
2) Have you checked if it works?

3. May 29, 2005

### dextercioby

Nope.It's a nonhomogenous I-st order linear equation with constant coefficients.Find an integrating factor.

Daniel.

4. May 29, 2005

### Icebreaker

Damn......

5. May 29, 2005

### whozum

I think it would be [itex] e^{2t} [/tex], right?

6. May 29, 2005

### arildno

That would be the integrating factor, yes.

7. May 29, 2005

### HallsofIvy

Staff Emeritus
Well, e2x, actually. $$\frac{d(e^{2x}y)}{dx}= 2e^{2x}y+ e^{2x}y'$$ so multiplying the entire equation by e2x gives e2xy'+ 2e2xy= (e2xy)'= (6x+5)e2x. Integrating the left side gives e2xy and the right side can be integrated by parts.

By the way, the reason your first method fails is that the integral of -2ydx is not -2xy because y is itself a function of x.