## ksaimouli find particular solution y=f(x one year ago one year ago

1. goformit100

its easy

2. ksaimouli

$\frac{ dy }{ dx }=6x^2-x^2y$

3. Spacelimbus

$\Large \frac{dy}{dx}=(6-y)x^2$ Or $\Large \frac{1}{6-y}dy=x^2dx$ For $$y(x) \neq 6$$

4. ksaimouli

$\int\limits_{}^{}\frac{ 1 }{ 6-y }dy=x^2dx$

5. Spacelimbus

Integrate both sides.

6. ksaimouli

$-lny=2x^3+c$

7. ksaimouli

@Spacelimbus

8. Spacelimbus

$\Large - \ln |6-y(x)|=\frac{1}{3}x^3+C\prime$

9. ksaimouli

they have given f(-1)=2

10. ksaimouli

no calc

11. ksaimouli

calculator

12. Spacelimbus

You know you have to solve it for y(x) if you can, of course - you could apply the initial conditions just now and solve for the constant. But I strongly recommend you to get the equation in explicit form if possible.

13. ksaimouli

i got $y(x)=ce ^{(x^3/x)}+6$

14. Spacelimbus

Almost, try again. You should end up with: $\Large y(x)=6-Ce^{-\frac{1}{3}x^3}$ As you can see, I said above that $$y(x) \neq 6$$ which was important, otherwise I would have divided the differential equation through 0. This equation supports that statement, this equation only becomes 6 in the limit as x approaches infinity.

15. ksaimouli

|dw:1360441531205:dw|

16. ksaimouli

|dw:1360441583641:dw|

17. ksaimouli

@Spacelimbus

18. ksaimouli

i alway get these

19. Spacelimbus

Consider this equation: $\Large e^{\ln2} = 2$ but: $\Large e^{-\ln2} = -2 ?????$ Rather view it like that: $\Large e^{-\ln2}=\left(e^{\ln2}\right)^{-1}=\frac{1}{2}$

20. ksaimouli

okay

21. Spacelimbus

and then you should get the same expression as I do.