Here is the question: y'' = (y')^2 (Assume y and y' are positive) I looked at the question and didn't know how to start it, never seen a question in this form. Help please!!!
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Originally Posted by Phyxius117 Here is the question: y'' = (y')^2 (Assume y and y' are positive) I looked at the question and didn't know how to start it, never seen a question in this form. Help please!!! Divide by and integrate once so Then separate and integrate again.
Thanks!! Now I got y = C1 e^(y) X + C2 as answer but how do I separate to just y in the equation with e^(y) present?
lol I think it's wrong but I plug y = lny' - C1 into y and becomes y= C1 e^(lny' - C1) + C2 which simplifies into y = C1y' + C2 then I integrate again becomes yx = C1y + C2x + C3 and finally y = (C2x + C3) / (X - C1) which I feel it's wrong...
Originally Posted by Phyxius117 lol I think it's wrong but I plug y = lny' - C1 into y and becomes y= C1 e^(lny' - C1) + C2 which simplifies into y = C1y' + C2 then I integrate again becomes yx = C1y + C2x + C3 and finally y = (C2x + C3) / (X - C1) which I feel it's wrong... Ya - you can't do that! Originally Posted by Danny Divide by and integrate once so Then separate and integrate again. From what I have. so (I absorbed the negative into the constants). Solving for y gives y .
Just put y'=f. The equation then becomes Separate and integrate to get and , Hence Now, Integrate to get the final answer
ty so much!!!
Note that Danny's answer is also correct: y where b=-ln(c1) and a=c2/c1
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