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# Maths Question watch

1. The question is:

Code:
```Find y in terms of x given that

dy/dx + 2y = sinx

and that the solution curve passes through the origin.```
I have got an answer which I think is right. But the answer in the back is different.
2. Maybe they're the same but rearranged... what have you got and what have they got?
3. (Original post by mikesgt2)
The question is:

Code:
```Find y in terms of x given that

dy/dx + 2y = sinx

and that the solution curve passes through the origin.```
I have got an answer which I think is right. But the answer in the back is different.
I = e^(integral 2 dx)
I = e^2x

e^2x dy/dx + 2e^2x y = e^2x sinx

e^2x y = integral e^2x sinx dx == K

u=e^2x => du/dx = 2e^2x
dv/dx = sinx => v=-cosx

K = -e^2x cosx + 2 {integral e^2x cosx dx == L}

a = e^2x => da/dx= 2 e^2x
db/dx = cosx => b=sinx

L = e^2x sinx - 2 integral e^2x sinx == e^2x sinx - 2K

K = -e^2x cosx + 2 e^2x sinx - 4K
5K = 2 e^2x sinx - e^2x cosx = e^2x (2six - cosx)

=> e^2x y = 1/5 e^2x (2sinx - cosx)

=> y = 1/5 (2sinx - 1osx + C)

0 = 2/5 sin 0 -1/5 sin 0 + 1/5 C

=> C = 1

=> y = 1/5 (2sinx - cosx +1)
4. (Original post by mikesgt2)
The question is:

Code:
```Find y in terms of x given that

dy/dx + 2y = sinx

and that the solution curve passes through the origin.```
I have got an answer which I think is right. But the answer in the back is different.
It's an integrating factor question with your IF being e^2x isn't it?
5. Elpaw...

I got what you got... apart from I just realised I made a stupid mistake in calculating the constant of integration.

But... the answer in the back is:

y = 1/5 ( 2sinx - cosx +e^(-2x) )
6. (Original post by mikesgt2)
Elpaw...

I got what you got... apart from I just realised I made a stupid mistake in calculating the constant of integration.

But... the answer in the back is:

y = 1/5 ( 2sinx - cosx +e^(-2x) )
yeah i se where that came from:

=> e^2x y = 1/5 e^2x (2sinx - cosx) + C

=> y = 1/5 (2sinx - cosx) + C e^-2x

dot dot dot
7. Oh rite, I see. Thanks for your help.

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