Katatonia

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4 years, 11 days

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These are replies submitted by Katatonia

@Rouben Rostamian  I really appreciate your kind help. Could you please let me know what were the problems? Should I add assuming t>0, t<T, T>t[1] always?

@dharr here, I have the final condition p(T)=0. 

I have uploaded the file and I would be thankful if you can help me

@acer Thank you so much

@nm Thank you so much

@Carl Love Thank you so much

 

@nm  Thank you. But, here you have assumed that g(x) is known. I just wanna have the form of the derivative. 

d( f(x)*g(x) ) /dx = f'(x)g(x) + g'(x)f(x). I wanna some thing like this, when g(x) is unknown. 

@nm Thanks for your comment. Isn't it possible to define it in Maple and take the derivative? I also want the expression for the derivative. In my case, the function is more complicated and I want to have something like 2g(x)+(2x+5)g'(x) for the first derivative and so on for higher-order derivative. 

 

Thanks

@Kitonum Thank you so much

@vv Thank you so much. So, at first, I should convert it to a polynomial and then, solve it. I want to find conditions which satisfy the following expression by I faced an error. 

 

-r+c+1 + sqrt(  r^2 - 2*r*c -r +2*c  ) >0,

It is the same the former case, but <1 is replaced with >0 with the same conditions on r and c. 

The error that I faced is "Error, (in SolveTools:-SemiAlgebraic) piecewise takes at least 2 parameters".

@Carl Love Thank you so much. Isn't it possible to have g(x)=max{h(x,p),f(x,p)} by conditioning on the value of x? For example, to represent g(x) as a piecewise function

@acer 

Thanks for your reply. In my problem, there are 6 variables but when I want to calculate the integration of the 6th variable, it does not calculate and just shows the formula of integration. I can compute the integral for Y0 to Y4 separately but when I want to calculate the integral for Y5, it just shows the formula.

For example, I want to find int(Y5(t)*lambda(t)*cosntant) where lambda(t) is a time-varying variable.

 

I wanna also add a constraint that Y0(t)+Y1(t)+Y2(0)+Y3(t)+Y4(t)+Y5(t)=1 for all t in the differential equation.

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