# Maths Question

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#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.
0
16 years ago
#2
Maybe they're the same but rearranged... what have you got and what have they got?
0
16 years ago
#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)
0
16 years ago
#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?
0
#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) )
0
16 years ago
#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
0
#7
Oh rite, I see. Thanks for your help.
0
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