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- Thread starter edgarpokemon
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Let's change y to t. On the right side you get an integral ## \int (t)(t+1)^3 \, dt ##.(I agree with your integrating factor of ## (y+1)^4 ##.) If you do it by parts using ## (1/4) \int t \, d(t+1)^4 ## , you do get the answer they give: ## x(t+1)^4=(1/4)t(t+1)^4-(1/4) \int (t+1)^4 \, dt =(1/4)t(t+1)^4-(1/20)(t+1)^5 +C ##. Finally, just divide out the ## (t+1)^4 ## and multiply by 20.

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- #5

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Let's change y to t. On the right side you get an integral ## \int (t)(t+1)^3 \, dt ##.(I agree with your integrating factor of ## (y+1)^4 ##.) If you do it by parts using ## (1/4) \int t \, d(t+1)^4 ## , you do get the answer they give.

but I have to do integration by parts on the right side right? So i get t(t+1)^4/4 - (t+1)^5/5=t(t+1)^4. I find a common multiple on the left side and I get a 20, and I divide (t+1)^4 on the right side and the left side and I get, 20t=5t-4(t+1)+c(t+1)^-4. which is 20t=t-4+c(t+1)^-4. but I am not sure what am i doing wrong. I now it has to do with some sign error but I can't find it.

- #6

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The "4" should not be there in front of (t+1). See my post #4 which I edited after you saw it. It's 20x= 5t-(t+1)+c(t+1)^{-4}. (There's a 1/4 factor in the second term of your first line.)but I have to do integration by parts on the right side right? So i get t(t+1)^4/4 - (t+1)^5/5=t(t+1)^4. I find a common multiple on the left side and I get a 20, and I divide (t+1)^4 on the right side and the left side and I get, 20t=5t-4(t+1)+c(t+1)^-4. which is 20t=t-4+c(t+1)^-4. but I am not sure what am i doing wrong. I now it has to do with some sign error but I can't find it.

- #7

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oooooohhh my God :'[ thank you for helping meee! engineering is not so for me I am changing majors! lolThe "4" should not be there in front of (t+1). See my post #4 which I edited after you saw it. It's 20x= 5t-(t+1)+c(t+1)^{-4}

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