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Consider the differential equation
dy 3x²
---- = ------
dx e²*

(because of my browser, the * is supposed to be a "y")

a) find a solution y = f(x) to the differential equation satisfying f(0) = 1/2

b) find the domain and range of the function "f" found in part a.

2007-03-13 19:14:27 · 2 answers · asked by Anonymous in Education & Reference Homework Help

2 answers

It's a separable differential equation, so if you "multiply" both sides by dx and e^(2y) and integrate, you get:

∫ e^(2y) dy = ∫ 3x² dx

Both sides of this are straightforward to integrate:

1/2 e^(2y) = x³ + C

Multiplying both sides by 2:

e^(2y) = 2x³ + C

Take the natural log of both sides:

2y = ln(2x³ + C)

And divide by 2 to solve for y:

f(x) = y = 1/2 ln(2x³ + C)

Then use the initial conditions to find C:

f(0) = 1/2 = 1/2 ln(2(0)³ + C)

1/2 = 1/2 ln(C)

1 = ln(C)

and then raising both sides to the e power:

e = C

So the function that satisfies the equation and initial condition should be:

y = f(x) = 1/2 ln(2x³ + e)

Just to double check, I'll differentiate:

y' = 1/2 (1/(2x³ + e)) (3x²) = 3x²/(2(2x³ + e))

but e^(2y) = e^(2(1/2) ln(2x³ + e)) = 2x³ + e

Somewhere there's a little mistake, but I can't find it, cuz, I end up getting:

y' = 3x²/(2e^(2y))

Which is wrong by a factor of 2 in the denominator...but I can't find my mistake...maybe you can.

b) If the correct function is:

f(x) = 1/2 ln(2x³ + e)

Then, for the domain, you need to ensure that the argument to the natural log does become zero or negative:

2x³ + e > 0
2x³ > -e
x³ > -e/2
x > ³√(-e/2)

So it appears the domain is (³√(-e/2), ∞).

The range of the function is the same as the range of natural log, (-∞, ∞).

2007-03-16 20:21:52 · answer #1 · answered by Jim Burnell 6 · 0 0

Integrating the function between x = 0 and x = 2 is perplexing as you're having to combine a logarithmic function. So, re-organize and combine in the different course. If y = 4ln(3 - x) then, y/4 = ln(3 - x) so, e^(y/4) = 3 - x => x = 3 - e^(y/4) Integrating w.r.t. y provides: 3y - 4e^(y/4) from y = 0 to y = 4ln3 Substituting provides: [12ln3 - 12] - [0 - 4] = 12ln3 - 8 Now, area R is the area of the rectangle minus the above subsequently: (6 x 2) - (12ln3 - 8) = 20 - 12ln3 = 6.817 Have a pass at something and allow me be attentive to in case you war......i like your philosophy on study techniques!! :)>

2016-12-14 18:41:50 · answer #2 · answered by claypoole 4 · 0 0

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