1st ODE !

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[x'=(x-t)^2+1\]
This thing isn't liniar any ideas ?
Hmm I have an idea.. but it's not giving me the same solution as Wolfram.. So I'm thinking I made a boo boo somewhere. I'll at least show you my attempt

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Let \(\large\rm u=x-t\) Differentiating our sub with respect to time gives \(\large\rm u'=x'-1\qquad\to\qquad x'=u'+1\) Subbing everything in gives us,\[\large\rm u'+1=(u)^2+1\]\[\large\rm u'=u^2\]And then just seperation, ya? :o
Ooo goodie! I actually am getting the same as wolfram, i just didn't simplify it far enough :)
Able to make sense of that? It should be correct :O
thx
|dw:1436763190196:dw| Am I right @zepdrix ?
A little hard to read that last line. Lemme just make a note that you should be careful to include your constant of integration at this step.\[\large\rm -\frac{1}{u}=t+c\]Put the negative on the other side,\[\large\rm \frac{1}{u}=c-t\]Solving for u,\[\large\rm \frac{1}{c-t}=u\]Then you need to undo your substitution. Remember, we're trying to solve for x(t).
aha thx again

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