consider the differential equation dy/dx=3x^2/e^2y

Question

(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)

Answer 1

Part A:

To find a solution to y = f(x) we start by separation of variables (this involves manipulating the differential equation so that expressions with x and y are on the same side as dx and dy, respectively.

dy/dx = 3x^2 / e^(2y)

e^(2y)dy = (3x^2)dx

∫e^(2y)dy = ∫(3x^2)dx

1/2e^(2y) = x^3 + C

To find the value of C, we substitute in the initial value f(0) = 1/2, in other words, (x,y) = (0,1/2)

1/2e^(2*1/2) = (0)^3 + C

1/2e = 0 + C

⇒ C = e/2

Using this value of C, we can write the particular solution y = f(x), solving for y

1/2e^(2y) = x^3 + e/2

e^(2y) = 2(x^3 + e/2)

e^(2y) = 2x^3 + e

Taking the natural log of both sides

ln(e^(2y)) = ln(2x^3 + e)

2y = ln(2x^3 + e)

y = 1/2*ln(2x^3 + e)

f(x) = 1/2*ln(2x^3 + e)

Part B:

To find the domain and range, we set the argument of natural log greater than zero and solve for x

2x^3 + e > 0

2x^3 > -e

x^3 > -e/2

x > (-e/2)^(1/3)

x > -1.108

Therefore the domain of f is

x ∈ (-1.108,+∞)

To find the range, we note that y = ln(x) has a range of

y ∈ (-∞,+∞)

Since y = ln(x) is the inverse of y = e^x, the domain of y = e^x becomes the range of y = ln(x). We know that y = e^x is defined for all real numbers.

Answer 2

Hello again Bethany!

Let me help out with this problem!

(a) Find a solution y = f(x)

Starting equation:

dy/dx = 3x² / e^2y

We start with a derivative, and want to integrate this and solve in terms of "y". Before we can try and integrate anything, we need to make sure that all of our "y-terms" are on the "dy" side, and all our "x-terms" are on the "dx" side...

dy/dx = 3x² / e^2y (multiply both sides by dx)

dy = [ 3x² / e^2y ] * dx (multiply both sides by e^2y)

[ e^2y ] dy = [ 3x² ] * dx

Now all the y's are on one side, and the x's are on the other side. We can now indefinitely integrate! Remember that indefinite integration always generates a +C integration constant!

∫ [ e^2y ] dy = ∫ [ 3x² ] * dx

(1/2)*e^2y = x³ + C

Now we just need to rearrange terms to solve in terms of y...

(1/2)*e^2y = x³ + C (multiply both sides by 2, note that 2*C is still an unknown arbitrary constant number, so we can still just call it +C for simplicity)

e^2y = 2x³ + C (take the natural log of both sides...)

ln(e^2y) = ln(2x³ + C) (ln() and exponent cancel on left side / apply log rule to right side...)

2y = ln(2x³ + C)

y = (1/2) * ln(2x³ + C)

What can we do about that integration constant C? Well let's plug in the 1 data point they told us...

f(0)=1/2

(1/2) = (1/2) * ln(2(0)³ + C)

1 = ln(C) (raise both sides e^)

C = e¹

C = e

So C = e in this situation as it turns out... Plug back in...

y = (1/2) * ln(2x³ + C)

y = (1/2) * ln(2x³ + e)

(b) What is the domain and range of this function?

This is a logarithm-based function. Logarithms have a range that spans from -∞ to +∞, so the range of this function is similarly...

Range: y∈(-∞ , +∞)

What about the domain? logarithms have a limited domain since you cannot plug negative (or zero) numbers into them. So, what values of x result in us plugging in an input greater than zero into the log?

2x³ + e > 0

2x³ > -e

x³ > -e/2

x > (-e/2)^(1/3)

So as long as our x-input is greater than (-e/2)^(1/3), the cube root of a negative half of Euler's number, we should be able to evaluate our logarithm numerically!

So the domain of our function can be written as...

Domain: x∈( (-e/2)^(1/3) , +∞)

Hope that helps!

--Zach

Answer 3