# More P3 IntegrationWatch

This discussion is closed.
#1
Find the general solutions of the differential equations in the following questions:

1) dy/dx = e^(2y-1).

2) dy/dx = cos^2 x.

3) dy/dx = cos^2 y.

4) dy/dx = xy.

5) dy/dx = e^(x+y).

1) 2x + e^(1-2y) = C.
2) 4y = 2x + sin2x + C.
3) tany = x + C.
4) 2lny = x^2 + C.
5) e^-y + e^x = C.

Thanks.
0
13 years ago
#2
I gotta go, so I'll give you some hints:

1. e^(1-2y) dy = dx
Use the substitution u=1-2y.

2. dy = cos²x dx
Use the identity cos²x = (cos2x + 1)/2

3. sec²y dy = dx
Recall that the integral of sec²y is tany.

4. (1/y) dy = x dx
The LHS will integrate to ln|y|, and the RHS is easy.

5. Use the rule: e^(x+y) = e^x . e^y, which will let you rewrite the equation as:
e^(-y) dy = e^x dx

I hope that helps.
0
13 years ago
#3
1) dy/dx = e^(2y-1).
dy/dx = e^(2y-1)
int e^(1-2y) dy= int dx
-1/2 e^(1-2y) = x + k
-e^(1-2y) = 2x - C
2x + e^(1-2y) = C

2) dy/dx = cos^2 x.
dy/dx = cos^2 x
int dy = int cos²x dx
cos²x = (cos2x + 1)/2
y = int (cos2x + 1)/2 dx
y = int [cos2x]/2 dx + int 1/2 dx
y = 1/4 . sin2x + x/2 + k
4y = sin2x + 2x + C

3) dy/dx = cos^2 y.
int 1/cos²y dy = int dx
x = int sec²y dy
x = tany - C
tany = x + C

4) dy/dx = xy.
dy/dx = xy
int 1/y dy = int x dx
ln y = x²/2 + k
2lny = x² + C

5) dy/dx = e^(x+y).
dy/dx = e^x . e^y
int e^-y dy = int e^x dx
-e^-y = e^x - C
e^x + e^-y = C
0
#4
(Original post by dvs)
I gotta go, so I'll give you some hints:

1. e^(1-2y) dy = dx
Use the substitution u=1-2y.

2. dy = cos²x dx
Use the identity cos²x = (cos2x - 1)/2

3. sec²y dy = dx
Recall that the integral of sec²y is tany.

4. (1/y) dy = x dx
The LHS will integrate to ln|y|, and the RHS is easy.

5. Use the rule: e^(x+y) = e^x . e^y, which will let you rewrite the equation as:
e^(-y) dy = e^x dx

I hope that helps.
Thanks for those.

I'm still struggling with the concepts of this - could anybody show the working out for a couple of these questions? I find it easier to understand this way. Cheers
0
#5
(Original post by yazan_l)
dy/dx = e^x . e^y
int e^-y dy = int e^x dx
-e^-y = e^x - C
e^x + e^-y = C
Why is "C" negative when integration has taken place?
0
13 years ago
#6
(Original post by Aristotle)
Why is "C" negative when integration has taken place?
Well it's not really. C can be anything.

I mean it's the same as the following:
∫ e^-y dy = ∫ e^x dx
-e^-y = e^x + c
e^x + e^-y = -c
let -c = C

Or the constant of integration could just be put on the LHS to begin with
∫ e^-y dy = ∫ e^x dx
-e^-y + C = e^x
e^x + e^-y = C
0
13 years ago
#7
C is any constant, i put it -ve so aas to be +ve when moving to the other side! i could had used :
-e^-y = e^x + c
-c = e^x + e^-y
C = e^x + e^-y
where C = -c
but to use less letters, which is meaningless, i put it -ve!
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