Simplify differential equation

[Han]

(02-05-2015 05:47 PM)lrdheat Wrote:  My TI NSpire and CASIO Classpad 400 come up with answers that agree with the book.

The TI answers (-2e^t^2)/((t^2-1)(e^t^2)-2c1)) I added ()'s to make this clearer.

The CASIO Classpad 400 produces a similar result.

According to the original post, the book solution is:
\[ y(t)=\frac{1}{ce^{t^2}-(2+t^2)} \]
If that is indeed what your book says, then it is wrong. You can check that against any CAS (even the TI).

Note that
\[ y(t) = \frac{-2e^{t^2}}{(t^2-1)e^{t^2}-2c1} \]
is the same as the HP Prime's
\[ y(t) = \frac{1}{\frac{1}{2}(-t^2+2G_0e^{-t^2}+1)} \]
by multiplying the the HP Prime's answer by \( \frac{-2e^{t^2}}{-2e^{t^2}} \)

parisse Wrote:lrdheat, the solution you give does not work unless I make a mistake
c:=-2*exp(t^2)/((t^2-1)*(exp(t^2)-2*c1));

It appears you put the parenthesis in the denominator around \( e^{t^2}-2c1 \) when it should only be on \( e^{t^2} \). lrdheat's post has mismatched ()'s

http://www.wolframalpha.com/input/?i=dso...*t*y%29%3B

Quote:The book suggest to put \( z=y^{(1-\alpha)} \) and in this case give for \( z'=-(2tz+t^{3}) \) the solution \( z(t)=ce^{t^{2}}-(2+t^{2}) \),

http://www.wolframalpha.com/input/?i=dso...2Bt%5E3%29