Hi, can someone please check my work. I only have one chance to submit my answer online and I wanna be 100% sure.

Results 1 to 3 of 3

- February 23rd 2014, 04:40 PM #1

- Joined
- Mar 2011
- Posts
- 22

- February 23rd 2014, 04:53 PM #2

- Joined
- Nov 2013
- From
- California
- Posts
- 2,789
- Thanks
- 1149

## Re: y' + 3y = sin(x)

this is a mess

are you solving for an integrating factor? (yes) then make it clear.

$$\mu(x)=e^{\int 3dx}=e^{3x}$$

then

$$\mu(x)\frac{dy}{dx}+\mu(x)y = \mu(x)\sin(x) \Rightarrow $$

$$e^{3x}\frac{dy}{dx}+e^{3x}y=e^{3x}\sin(x)$$

$$\frac{d}{dx}\left(e^{3x}y\right)=e^{3x}\sin(x)$$

etc.

Then, you properly introduced the integration constant C, but then lost it.

You need to include that and solve for it given your initial condition.

Rework it.

- February 23rd 2014, 06:08 PM #3

- Joined
- Mar 2011
- Posts
- 22