## ksaimouli 2 years ago solve

1. ksaimouli

$\frac{ dy }{ dx }=x^2e^{4x}$

2. ksaimouli

solve the differential equation

3. ksaimouli

i tried using by parts

4. satellite73

i think that just means find the anti derivative of $$x^2e^{4x}$$and my guess is use parts twice

5. ksaimouli

ya i used

6. ksaimouli

but not solved let me show

7. ksaimouli

i will use 7 technique

8. ksaimouli

|dw:1360895129811:dw|

9. ksaimouli

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10. ksaimouli

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11. satellite73

i think you have it backwards you want to drop the power on the $$x^2$$ term

12. ksaimouli

and now i should choose x^3 and e^4x

13. satellite73

no the gimmick is to reduce the power, not raise it. raising the power makes it worse

14. satellite73

$$u=x^2, dv = e^{4x}dx, du = 2x, v =\frac{1}{4}e^{4x}$$

15. ksaimouli

okay i can do it this way so something wrong in my technique right

16. satellite73

yes. as you can see your method just gave higher and higher powers of $$x$$ you want to make them lower and lower

17. ksaimouli

so instead of integrating id i take derivative

18. ksaimouli

but if i follow the LIPET rule it says u= log right

19. satellite73

i forget exactly how that box method works, but it is $\int u dv =uv-\int v du$

20. satellite73

make $$u=x^2, du = 2xdx, dv = e^{4x}dx, v =\frac{1}{4}e^{4x}$$ get $\int x^2e^{4x}=\frac{1}{4}e^{4x}x^2-\frac{1}{2}\int e^{4x}xdx$ then do it again

21. ksaimouli

okay got u

22. ksaimouli

thx

23. satellite73

yw