Recent content by coomast

Hello juice34, It does not seem that difficult. I assume that the independent variable is r and that the dependent one is v. The first step is to substitute the second equation into the first one and leaving out the part which is divided away with regards to the zero on the right hand side...

CremeBrulee, I do not know what exactly is meant by the cover-up method, but these two examples are very straightforward. You start very well, but than you seem to be lost. I will explain the first one, the second one you need to do yourself. (I wrote the first one down on a piece of paper in...

The left hand side and the right hand side must be equal, thus you need to set the factors for the s term and the constant equal to each other. This gives a (very easy) system of two equations in two unknowns (a and b): 1=a 4=2a+b I assume you can solve this. Don't make it too difficult...

Nice Cody Palmer, very nice :approve: coomast

Hello Leo Klem, Not an easy one. The only method I know that can be used to solve this equation is Lie point transformations. By setting: u=y v=y' and w=\frac{dv}{du}=\frac{y''}{y'} You can transform the equation into: v\frac{dv}{du}=\frac{b}{u^2}-au Which has as solution...

No, it isn't according to my calculations. I got: \frac{3}{x}\cdot e^x I checked by putting the complete solution y=\left(K_1 \cdot x + K_2 +\frac{3}{x}\right) \cdot e^x back into the original DE. coomast

This is the same result I had. Now the initial remark raised by CRGreathouse was: "This can be solved fairly quickly compared to a general Newton's method based solution." What is this other method? Using Newton-Raphson on the Lambert function gives in the end the same interative scheme as the...

Hello qraal, If the x-values are larger than 2 it has a unique solution. In case these can be between 1 an 2 as well you will have two solutions. This is clear from the picture and a substitution in the equation. Indeed, we have for x=1 y(1)=\frac{1}{ln(2)} whichis the same value for...

Hello qraal, Using the method of Newton-Raphson will give you a fast approximate solution if you have a "good" starting value in the neighbourhood of the solution. The method can be found here: http://en.wikipedia.org/wiki/Newton's_method and is for your equation...

Mmmmm, I get the following: \alpha \cdot x + \beta = \int \kappa(T) \cdot dT In which \alpha and \beta are the integration constants. You can further calculate the integral once you know the dependency of the conductivity with temperature. Does this clearify things? coomast

Hello mherna48, Since A is a constant it is divided away with regards to the right hand side. We are left with a derivative equal to 0, giving thus a constant as solution. This means we have (ordinary derivatives because only x is the independent variable and only x): \kappa(T) \cdot...

Though not the question it is interesting to know that this DE has a solution in closed form. One can find this by substituting x=t^2 in the equation. A very simple DE will appear and can be solved directly. Doing the inverse substituting on this solution gives the result of the original DE...

It is a partial differential equation, both t and x are the independent variables and P the dependent function to find. C is a given function, but I don't see how to solve it. Certainly numerical ways should be possible but I can't help you on this. coomast

pfff...the preview gives a different formula then the one I'm typing, I assume the latex is still broken This is done in the same way as before, you have: v\frac{\partial u}{\partial y}=v^*U\frac{\partial u^*U}{\partial y^*L} = \frac{U^2}{L} v^*\frac{\partial u^*}{\partial y^*} After...

I had a look at it and after a few minutes I got a transformed result, but indeed it is not the one given. However there seems to be a term missing in the left hand side containing the time t. After using the transformation I got (without the term with the time): u^*\frac{\partial...