1. ## differential equations

I have a whole set of differential equations that I have to solve, so I am going to list them all, but if someone could just help me with one or two of them, I should be able to solve the rest??? Thanks.

1. xy' = y + 1

2. y" + y' - 2y = sin x + e^x

3. y" + y = sin x, y(0) = 1, y'(0) = 0

4. y"' - y' = xe^x

5. (1 - x^2)y" - 2xy' + 42y = 0

(number 4 looks different than the others, so maybe if someone could help me with that one and one other I would be able to solve the rest. Thank you very much!)

2. Hello, SoBeautiful!

The first one is separable . . .

. . . . dy
1) .x·---- .= .y + 1
. . . . dx

. . . . . . . . . .dy . . . .dx
We have: . ------- .= .---
. . . . . . . . .y + 1 . . . x

Integrate: .ln|y + 1| .= .ln|x| + c .= .ln|Cx|

Therefore: .y + 1 .= .Cx . . . . y .= .Cx - 1

3. Thanks, that was helpful... I feel silly for not seeing it. But I am still not sure about what to do with the others, so if someone could help me with one more, that would be amazing. Thank you!

4. Originally Posted by SoBeautiful
2. y" + y' - 2y = sin x + e^x
This is a linear constant coefficient inhomogeneous equation

So the general solution is the sum of the general solution of the
homogeneous equation:

y" + y' - 2y = 0

and a paticular integral of the DE.

The general solution of the homogeneous equation is obtained by considering
solutions of the form y=e^{lambda x}, then:

lambda^2 + lambda -2 = 0

is the indicial equation, and so lambda = 2 or -1 and the general solution of
the homogeneous equation is:

y = Ae^{-2x} + Be^{x}.

For a particular integral consider a solution of the form:

y= u sin(x) + v cos(x) + w x e^{x}

then:

y' = u cos(x) - v sin(x) + w[xe^{x} + e^{x}]

y'' = -u sin(x) - v cos(x) + w[xe^{x} + 2e^{x}]

Substituting these into the DE gives:

sin(x)[-u - v - 2u] + cos(x)[-v + u -2v] + 3we^{x}=sin(x) + e^{x}

So w=1/3, and u=-3/10 and v=-1/10.

So the general solution of the original DE is:

y = Ae^{-2x} + Be^{x} - (3/10)sin(x) - (1/10)cos(x) + (1/3)x e^{x}

RonL

5. Originally Posted by SoBeautiful

5. (1 - x^2)y" - 2xy' + 42y = 0

Here's my partial attempt at number 5. hopefully you can finish it, and hopefully I haven't made a mistake somewhere

(1 - x^2)y" - 2xy' + 42y = 0

The corresponding indicial equation is:
To find solution of the form, y = x^r, we get,

-r(r - 1) - 2r + 42 = 0
=> -r^2 + r - 2r + 42 = 0
=> r^2 + r - 42 = 0
=> (r + 7)(r - 6) = 0
=> r = -7, r = 6

Take the highest power, that is r = 6
Now we must find the series solution of the form, y = SUM{0^inf}a_n*x^(n + 6)

6. Originally Posted by SoBeautiful

4. y"' - y' = xe^x
Reduction of order, define u=y'

7. Originally Posted by SoBeautiful
5. (1 - x^2)y" - 2xy' + 42y = 0
This can be more appropriatly solved with Legendre Polynomials.

This equation happens to have polynomial solutions.

8. Originally Posted by ThePerfectHacker
This can be more appropriatly solved with Legendre Polynomials.

This equation happens to have polynomial solutions.
that looks familiar. i think i saw it in my text, but i don't recall doing it in class

9. Originally Posted by Jhevon
that looks familiar. i think i saw it in my text, but i don't recall doing it in class
These are not done in class.

You learn these types of things in a mathematical physics course. They are part of "special functions" (those there is nothing "special" about Legendre polynomials). People who study Quatum Mechanics probably come across them.

They also appear when you consider the 3 dimensional Laplace equation and you convert to Spherical Coordinates.