# Math Help - General Solution to an Equation

1. ## General Solution to an Equation

Hi,

I was wondering if someone could take five minutes to explain how you work out the general solution to an equation. I sort of understand it but is there a way i can look at an equation and instantly recognise that a particular equation has a certain solution. For instance:

dy/dx = 3(x -1)^2 + 20(y-(x-1)^3)

has the general solution:

y(x) = (x-1)^3 + Ae^20x

How would i go about finding that?

Does it depend on then inital conditions and the boundary conditions?

Cheers for the help.

2. "Five minutes"?

That is an extremely big question.

Not all differential equations have solutions that can be determined analytically. In fact the vast majority can't be.

The ones you're likely to get in an introductory course will be carefully taylored (deliberate pun) to fit the techniques which you will be guided through one by one. Some d.e's are solvable using easy techniques, some by more difficult ones.

This one:

$\frac {dy}{dx} = 3(x-1)^2 + 20 (y-(x-1)^3)$

looks at first glance to lend itself to an integrating factor approach, but this is an area in which I'm rusty and would need to check. You may be able to separate the variables.

Your five minutes are up, I'm afraid.

3. $\frac{dy}{dx}-20y=f(x)$

$\text{i.f.}=e^{\int -20 dx}=e^{-20x}$

$\int \left(e^{-20x}y\right)'=\int e^{-20x}f(x)$

$y=e^{20x}\left[\int e^{-20x}\left(3(x-1)^2-20(x-1)^3\right)\right]$

$y=e^{20x}\left[e^{-20x}(x-1)^3+c\right]$

$y=(x-1)^3+e^{20x}c$

Guess that might be a bit cryptic. Lots of books skip steps. They're meant to encourage you to fill in the blanks and that helps you learn it.