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How can wolfram alpha be used to find the solution to an IVP, if the IVP has non integers in the condition For example, enter image description here The half in the function was changed to multiply outside(it does this for any noninteger) enter image description here But as shown above works for any integer.

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jamie
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    Just change your input: $a'(b)=2*b / a^3, a(0.5)=1$ You always have to check how your input is interpreted. – callculus42 Jan 21 '20 at 18:01
  • The problem is that in both other cases where you used $a$, you used it without an argument - its dependence on $b$ was implicit. WA assumed that your final use of $a$ was the same, and the (0.5) was a multiplier, in parentheses because of the radix. In addition to callculus's method, you can make every reference to $a$ be explicit in its dependence on $b$: da(b)/db = 2*b/a(b)^3, a(.5) = 1 – Paul Sinclair Jan 22 '20 at 01:23

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