General Solution to Linear, Homogenius, Second-Order Differential Equation

[nivekious]

I need to solve d/dx(e^(-(x^2))*(du/dx))=u+x*(du/dx)
I get e^(-(x^2))*u''-2xe^(-(x^2))*u'=u+x*u'
u''-2x*u'=e^(x^2)*u+xe^(x^2)*u'
u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0
but I have no idea how to approach the problem from here. Can somebody please help?

 

[jackmell]

I need to solve d/dx(e^(-(x^2))*(du/dx))=u+x*(du/dx)
I get e^(-(x^2))*u''-2xe^(-(x^2))*u'=u+x*u'
u''-2x*u'=e^(x^2)*u+xe^(x^2)*u'
u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0
but I have no idea how to approach the problem from here. Can somebody please help?

I can suggest a route to study it. First, get a DE text that deals with second-order DEs with variable coefficients. I use "Intermediate Differential Equations" by Rainville. There are some simple substitutions to try with these types of equations, putting it in it's normal form $v''+Iv=0$ and sometimes that results in a constant-coefficient equation. Some other substitution method to get a constant-coefficient equation. I tried those quickly and did not seem to do it. But you may want to try yourself. My next step would be to turn to Mathematica in desperation using the DSolve command. I tried that and Mathematica can't solve it. My next step, even more desperate but not too uncommon is to use power series which of course in this case will involve Cauchy-products but that's ok according to Rainville. If I had to solve it analytically, and I could think of nothing else, I'd do it with power series.

 

[JJacquelin]

General Solution to Linear, Homogenius, Second-Order Differential Equation

I don't think that the ODE is on the homogenius kind. It doesn't matter since direct integration directly leads to a first order ODE :

 

[jackmell]

. . . schooled. Very nice though. :)

 

[blue_raver22]

I can provide some guidance on how to approach this problem.

Firstly, it is important to understand the characteristics of the given differential equation. It is a linear, homogeneous, and second-order equation. This means that the dependent variable u and its derivatives are only raised to the first power, there are no non-linear terms, and the equation is free of any constants.

To solve this type of equation, we can use the method of variation of parameters. This involves finding a particular solution and then using it to find the general solution.

To find a particular solution, we can assume that u = Ae^(kx) where A is a constant and k is a constant to be determined. Substituting this into the equation, we get:

Ae^(kx)*(-2kx-2x^2) - Ae^(kx)*(2x-xe^(x^2))k - Ae^(kx)*e^(x^2) = 0

Simplifying, we get:

Ae^(kx)*(-2kx-2x^2+2kx-kx*e^(x^2)-e^(x^2)) = 0

Since e^(x^2) cannot equal 0, we can equate the expression inside the parentheses to 0. This gives us the following equation:

-2kx-2x^2+2kx-kx*e^(x^2)-e^(x^2) = 0

We can solve for k and get k = 1. Substituting this back into our assumed solution, we get:

u = Ae^x

Now, we can use this particular solution to find the general solution. The general solution will have two parts - the particular solution and the complementary function. The complementary function is the solution to the homogeneous equation, which in this case is:

u''+(2x-xe^(x^2))*u'-e^(x^2)*u=0

This can be solved using the method of undetermined coefficients or by using the characteristic equation. The complementary function is given by:

u = Be^(x^2)

Hence, the general solution is:

u = Ae^x + Be^(x^2)

In conclusion, the method of variation of parameters can be used to solve this type of differential equation. By finding a particular solution and combining it with the complementary function,