Homework Help: Another First-Order ODE Question

1. Jan 6, 2009

bsodmike

1. The problem statement, all variables and given/known data

Going through a book on Numerical Methods, it states,

$$y' = 4e^{0.8x}-0.5y$$

has the analytical solution,

$$y= \dfrac{4}{1.3}(e^{0.8x}-e^{-0.5x})+2e^{-0.5x}$$

2. Relevant equations

This is of the form,

$$y'+p(x)y=q(x)$$

Should I use an Integrating Factor to solve the Linear ODE?, i.e. use an integrating factor $$\mu(x)=e^{\int{p(x)dx}}$$

$$\mu(x)\left[y'+p(x)y\right]=\mu(x)q(x)$$

$$\left(\mu(x)y\right)'=\mu(x)q(x)$$

$$\mu(x)y=\int{\mu(x)q(x)dx+C}$$

...and divide through by $$\mu(x)$$ ???

3. The attempt at a solution

$$\mu(x)=e^{\int{p(x)dx}}=e^{0.5x}$$

thus,
$$e^{0.5x}y=4\int{e^{0.8x}e^{0.5x}dx+C}$$

and it is given: y(0) = 2

Any pointers? Dick, your help would be much appreciated once again!

Thanks and happy new year to all as well,

Cheers
Mike

Last edited: Jan 6, 2009
2. Jan 6, 2009

Integral

Staff Emeritus
When computing your integrating factor what did you use as your p(t)? Your integrating factor is not correct.

3. Jan 6, 2009

bsodmike

Note: with p(t)dt, consider it as p(x)dx. I used some old latex definition I had already typed for the 'method'. i.e., rather than dy/dt, this is dy,dx. dy/dt = f(t,y) -> dy/dx = f(x,y).

*Definition updated in OP, w.r.t to (x).

Last edited: Jan 6, 2009
4. Jan 6, 2009

bsodmike

I used,

$$\mu(x) = e^{\int{p(x)dx}}=e^{\int{\dfrac{1}{2}dx}} = e^{0.5x}$$

Well, isn't this,

$$y'+0.5y=4e^{0.8x}=y'+p(x)y=q(x)$$

hence, $$p(x)=0.5$$ and $$q(x)=4e^{0.8x}$$ ???

Last edited: Jan 6, 2009
5. Jan 6, 2009

Integral

Staff Emeritus
$$e^{\int{\dfrac{1}{2}dx}}$$ Try doing this integral again. Where did the ln 2 come from?

6. Jan 6, 2009

bsodmike

Oooops!!! thanks for spotting that blatant error. $$IF=e^{\int{p(x)dx}}=e^{0.5x}$$.

7. Jan 6, 2009

bsodmike

This is what I am getting now,

Since,
$$\mu(x)=e^{\int{0.5\,dx}}=e^{0.5x}$$

$$\begin{split} e^{0.5x}y&=4\int{e^{1.3x}}\,dx+C\\ &=4\left(\dfrac{e^{1.3x}}{1.3}\right)+C\\ &=\dfrac{4}{1.3}e^{1.3x}+C \label{eq:} \end{split}$$

Hence,
$$\begin{split} y=\dfrac{4}{1.3}e^{0.8x}+Ce^{-0.5x} \label{eq:} \end{split}$$

Where am I going wrong again?!?

Last edited: Jan 7, 2009
8. Jan 7, 2009

bsodmike

*bump*

When I left the problem earlier as $$e^{0.5x}y=4\int{e^{0.8x}e^{0.5x}\,dx}+C$$, I thought I'd need to use integration by parts. But once I got it in the simpler form above (adding the powers) it just seems 'too simple' and the solution is missing a term.

*confused*...hmm..