## anonymous one year ago Eliminate the arbitrary constants. Any solutions. Problem is given by the picture.

• This Question is Open
1. anonymous

2. Haseeb96

is there any choices?

3. Haseeb96

Well, there is no any solution to eliminate those constants. Because both are different

4. UsukiDoll

something's off. it's like the general solution for a second order ode with a repeated root.

5. UsukiDoll

well I can't stay up... I'm off to bed.

6. anonymous

the book say the answer is y"-2y'+y=0 i do not know how to solve it.

7. anonymous

but the problem is to eleminate the arbitrary constants which are c1 and c2.

8. imqwerty

all u have to do is differentiation :) hope this helps

9. anonymous

but how? I am having trouble of iy, i know how to get the derivative but i do not know how to eliminate.

10. freckles

@njwild you keep saying you want to eliminate the arbitrary constants but you said the answer was a differential equation so it makes me think you want to work from y=c1e^x+c2xe^x to get y''-2y'+y=0 . But if you are looking for the constants than you initial conditions.

11. anonymous

but how to solve it by a given problem to get the answer with solution.

12. imqwerty

can u find the value of y' and y' by differentiation??

13. anonymous

after differentiation imqwerty , what is the steps of rliminating or the first to eliminate until last to get the answer

14. anonymous

@imqwerty I get the y' and y" but i do not know what is the first to eliminate until last to get the answer.

15. imqwerty

wait m tellin :)

16. imqwerty

it would take too long to type btw heres the solution - https://www.youtube.com/watch?v=HIdKpnWb2Ws

17. imqwerty

nt exactly the same but it may help u

18. freckles

If you are trying to get the differential equation from the solution here is my explanation: $y=c_1e^x+c_2xe^{x} \\ \text{ the } x \text{ next to the other } e^x \text{ tells us we have a repeated solution \to the } \\ \text{ I will call it the characteristic equation } \\ \text{ since the solution is } y=c_1e^{1 \cdot x}+c_2 xe^{1 \cdot x} \\ \text{ then the solution to the charateristic equation } is r=1 \\ r-1=0 \\ (r-1)^2=0 \text{ since the solution is repeated } \\ r^2-2r+1=0$ from here it should be easy to write the differential equation

19. freckles

but to figure out what the constants are you need conditions such as y(a)=b and y'(c)=d whatever a,b,c, and d are given as

20. freckles

other than that I have no idea what you mean by eliminate the arbitrary constants

21. UsukiDoll

@freckles that's what I had typed earlier because we had to look for the repeated root, but it was getting late and the asker drove me nuts, so I was erasing my posts before I logged off. Users should stop posting generic questions and add more details instead of one liners.

22. anonymous

differentiate twice and then eliminate c1 and c2 from three equations.