what is the solution of the initial value problem?

Question

dy/dx= - square root of 16y^3 e^4x (y>0), y=1/4 when x=0

Answer 1

i guess the sqrt is for the entire expression....

rewrite it as:

-dy/y^(3/2) = 4 e^2x dx (after applying sqrt)

integrating both sides, we get

2/y^(1/2) = 4 e^2x / 2 + c

or 2/y^(1/2) = 2 e^2x +c

putting x=0, y=1/4 we can get the value of the constant, c

2/(1/2) = 2 + c or c = 4-2 = 2

hence the solution is 2/y^(1/2) = 2 e^2x + 2

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

Answer 2

Finally, something to sink my teeth into. This equation is a "seperable equation" because it can be written in the form
dy/dx = g(x) f(y)
or, equivalently dy/dx = g(x)/h(y)

We write this as
dy/dx = sqrrt(16 exp{4x}) / - sqrrt (y^-3)

This can be reordered as
- sqrrt(y^-3) dy = sqrrt (16 exp{4x}) dx
or
- y^(-3/2) dy = 4 exp{2x} dx

Integrating the left hand side gives
- (-1/2)^(-1) y^(-1/2) = - (-2) y^(-1/2) = 2y^(-1/2)

Integrating the right hand side gives
4 (1/2) exp{2x} + K = 2 exp{2x} + K

(where K is a constant)

So
2y^(-1/2) = 2 exp{2x} + K
y^(-1/2) = exp{2x} + 1/2 K
y^(-1) = [exp{2x} + 1/2 K]^2
y = [exp{2x} + 1/2 K]^(-2) (*)

Setting x = 0 and y = 1/4,
1/4 = [exp{0} + 1/2 K]^(-2)
4 = [1+ 1/2 K]^2
[1+ 1/2 K] = 2 OR [1+ 1/2 K] = -2 (taking the squareroot of 4 as 2 or -2)
1/2 K = 1 OR 1/2 K = -3
K = 2 OR K = -6

From (*), this gives
y = 1 / ((exp{2x}+1)^2)
or
y = 1 / ((exp{2x}-3)^2)

You can test both of these by taking the derivative with respect to x and you'll get the correct answer - I'll just quickly show you with the first one:

dy/dx = d/dx [exp{2x}+1]^(-2)
= (-2) * [exp{2x}+1]^(-3) * (2) * exp{2x}

Now, y^(1/2) = [exp{2x}+1]^(-1)

so dy/dx = -4 * y^(-3/2) * exp{2x}

which leads back to the original equation giving, so the solution works.

That ok?

Answer 3

6

Answer 4

In mathematics, an initial value problem is a statement of a differential equation together with specified value of the unknown function at a given point in the domain of the solution. Calling the given point t0 and the specified value y0, the initial value problem is


The problem is then to determine the function y.

This statement subsumes problems of higher order, by interpreting y as a vector. For derivatives of second or higher order, new variables (elements of the vector y) are introduced.

More generally, the unknown function y can take values on infinite dimensional spaces, such as Banach spaces or spaces of distributions.

Answer 5

Your question has no known answer as the square root of 16y^3 is under the value of y2 and is there for more than the ratio of y which would = 14 of more...: Thus rendering your question a statement of error rather than a answer seeker.