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I know that it's possible to get the step by step solution of a differential equation calling Wolfram in Mathematica. For example:

WolframAlpha["L*x'+R*x=Sqrt[(sin(t))^2] and x(0)=0", 
 PodStates -> {"Step-by-step solution"}]

I tried to do the same thing with another differential equation, but it does not work:

WolframAlpha["Qa Ca'(x)=ka Ca(x) and Ca(0)=Ca0", 
 PodStates -> {"Step-by-step solution"}]

How can I get the step by step solution?

Thank you so much for your time.

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    $\begingroup$ Your variable names seem to be confusing Alpha. Try this: WolframAlpha[ "q*(c'[x])==k*c[x]&&c[0]==c0", PodStates -> {"Step-by-step solution"} ] $\endgroup$ – b3m2a1 Nov 15 '17 at 8:48
  • $\begingroup$ Hello @b3m2a1. Thank you for your good solution. For me it's important to preserve the original notation because, when I solve a system of differential equations, the changing of the notation makes the solution obscure. $\endgroup$ – Gennaro Arguzzi Nov 15 '17 at 9:01
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Ca is a token for Wolfram Alpha. It is from the chemical elements, i.e. calcium. It can not be taken anymore as a variable or function or constant for symbolic knowledge computation.

It is not understandable for Wolfram Alpha what f(0)=f0 is intended for. It always interprets this input as another defining equation, not as a boundary value input. So there is a need for using another constant name.

Factors or constants are desired to have single-letter names otherwise they an interpreted

So this input works already:

WolframAlpha["q*c'(x)=k*c(x) and c(0)=c0, PodStates -> {"Step-by-step solution"}]

The output on the internet is

Input: {q c'(x) = k c(x), c(0) = c0} Separable equation: (c'(x))/(c(x)/q) = k ODE classification: first-order linear ordinary differential equation Alternate form: {k c(x) = q c'(x), c0 = c(0)} Differential equation solution: c(x) = c0 e^((k x)/q)

The typing is not really accurate by Wolfram Alpha. It names it a first-order linear ordinary differential equation, but it is handled as a first-order linear ordinary differential equation with constant coefficients and a single boundary condition.

You can google for that type of differential equation, look up in collections, libraries, books that a reliable. It is typed a lit sluggish but handled correctly by Wolfram Alpha and it is a homogeneous variant contrary to the given example that is inhomogenous by the square root of the sine squared.

There need for an integration constant for the first order of the differential equation. You are free to name your boundary value symbolically as long as this is not the name of the function of the differential solution or the domain variable.

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  • $\begingroup$ Hello @user2432923 thank you a lot for your clear reply. $\endgroup$ – Gennaro Arguzzi Mar 27 '20 at 9:50

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