Linear Differential Equation

Question

So I asked this question a while back but the answer I recorded really did not help me solve it.

Find the general solution to:

$a_0 + a_1x + a_2y + a_3dy/dx = 0$

If $a_0 = 0$ this becomes quite easy to solve (just set y = vx and factor it) but separation of variables doesn't work otherwise.

I think it is important to note that this is non homogenous.

Can someone give the general solution with an explanation of how it was solved (and if it isn't too much) as well as an explanation of intuitions/motivations behind the methods used?

My phone screwed up... I corrected the tags. I was planning on posting a different question but had a certain change of heart — Sidharth Ghoshal, Jun 24 '13 at 04:51
@mhenni I should thank you and say that ure comment was half the answer I was looking for — Sidharth Ghoshal, Jun 24 '13 at 05:10

score 3 · Accepted Answer · answered Jun 24 '13 at 05:00

3

Hints:

Rewrite it as $\displaystyle y' + \frac{a_2}{a_3} y = -\frac{a_0}{a_3} - \frac{a_1}{a_3} x$
Find an integrating factor

After those steps, you'll end up with:

$$\displaystyle \large y(x) = -\frac{a_0}{a_2}+\frac{a_1 a_3}{a_2^2}-\frac{a_1 x}{a_2}+c_1 e^{-\frac{a_2 x}{a_3}}$$

As an alternate approach, you could solve the homogeneous equation and then guess at a particular solution and find the constants.

answered Jun 24 '13 at 05:00

Amzoti

56,093

Let me start working it out – Sidharth Ghoshal Jun 24 '13 at 05:02
How exactly did u find the integration factor? I take it that it is some functional expression that must be multiplied to both sides to make it into a solvable form – Sidharth Ghoshal Jun 24 '13 at 05:04
@frogeyedpeas: Are you actually familiar with IF? – Amzoti Jun 24 '13 at 05:05
Not really... To be honest – Sidharth Ghoshal Jun 24 '13 at 05:06
I read the technique on the comments this makes more sense now! – Sidharth Ghoshal Jun 24 '13 at 05:10
@frogeyedpeas: Excellent, have fun! Regards – Amzoti Jun 24 '13 at 05:11
Nice work, again, my friend! + 1 – amWhy Jun 24 '13 at 16:56
Yes, I've got to attend to some actual work, too ;-) – amWhy Jun 24 '13 at 17:01

Linear Differential Equation

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