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[✓] Solve a normal Differental-Equation?

Karl Eberl
Posted 5 years ago

Hello Everyone, I am trying to learn Mathematica with a Book, but I have a Problem to Solve an Equation, although I have a AnswerKey of the Examples.

In General i have to solve the equation y''[x]+y'[x]==x²*Sqrt[x],y[x],x]. First i want to solve it with C1 and C2, later i want to replace them and plot the function. Well, I think my ProbleM is the first Part - just solve it with C1 and C2. My Code is: 

DSolveValue[y''[x] + y'[x] == x^2*Sqrt[x], y[x], x]`

The Code is exactly the same as in the AnswerKey to this Example, but as a solution I get:

(2 x^(7/2))/7 - E^-x C[1] + C[2] + ( E^-x Sqrt[-x] Gamma[7/2, -x])/Sqrt[x]

the "Right" solution from the answer Key should be:

C[2] + 1/56 E^-x*(2*(E^x*Sqrt[x]*(-105 + 70 x - 28 x^2 + 8 x^3) - 
28*C[1]) + 105*Sqrt[Pi]*Erfi[Sqrt[x]])

So the most obvious difference for me is that "Erfi" function in die Solution and in my solution there is a "Gamma" function. I already made some Research About them but didn't really understood what they do. Furthermore, I don't get why my solution differ from the Answerkey one, although I have the same Input. In the Attachments is my Mathematica file (it is the first Input) and a PDF of the AnswerKey how the solution should look like. I hope someone can tell me what to do.

Attachments:
Problem Chapter 16.nb
Solution 4.pdf
POSTED BY: Karl Eberl
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Karl Eberl
Karl Eberl
Posted 5 years ago

Hello Mr. Iwaniuk, Thanks for your Reply :)

I tried it, and have the same Solution than you. but if i replace the C1 and C2, and want to plot it, it seems that the Solution is different to the Solution from the Answerkey. If i Plot with the Code of your solution, i need to plot from {x,0,100} to see the graph. In the solution it should start from ~0 In the Attachments are the Solutions from the Answerkey again :)

Do you have any idea whats going on there? Thanks for your help.

Attachment

Attachments:
Chatper 16 - Exercises.nb
Solution Problem.JPG
POSTED BY: Karl Eberl

It's seems both solution are correct only for Real numbers. See attached file.

Attachments:
Chatper 16 - Exe...nb
POSTED BY: Mariusz Iwaniuk
Karl Eberl
Karl Eberl
Posted 5 years ago

Ah okay, I never would have thought about this. Thanks a Lot for your help :) I really appreciate it.
POSTED BY: Karl Eberl

Try:

 eq = y''[x] + y'[x] == x^2*Sqrt[x];
 sol = DSolve[eq, y, x]
 (* {{y -> Function[{x}, (2 x^(7/2))/7 - E^-x C[1] + C[2] + (
      E^-x Sqrt[-x] Gamma[7/2, -x])/Sqrt[x]]}} *)
 eq /. sol // FullSimplify(*Solution is Correct*)
 (* {True}*)

Using FunctionExpand to get Erfi[x] function:

 sol2 = DSolveValue[eq, y[x], x] // FunctionExpand // Simplify
(*-((15 Sqrt[x])/4) + (5 x^(3/2))/2 - x^(5/2) + (2 x^(7/2))/7 + 
 E^-x ((15 Sqrt[\[Pi]] Sqrt[-x])/(8 Sqrt[x]) - C[1]) + C[2] + 
 15/8 E^-x Sqrt[\[Pi]] Erfi[Sqrt[x]] *)

eq /. {{y -> Function[{x}, Evaluate@sol2]}} // FullSimplify(*Solution is correct*)
  (* {True}*)

Check if yours solution are correct:

eq /. {{y -> 
      Function[{x}, 
       C[2] + 1/
          56 E^-x (2 (E^x Sqrt[x] (-105 + 70 x - 28 x^2 + 8 x^3) - 
              28 C[1]) + 105 Sqrt[Pi]*Erfi[Sqrt[x]])]}} // FullSimplify
(* {True}*)

Regards M.I.
POSTED BY: Mariusz Iwaniuk
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