Consider the differential equation dy/dt = y or sometimes written y\' - y. For a
ID: 2882302 • Letter: C
Question
Consider the differential equation dy/dt = y or sometimes written y' - y. For any constant C, the function y(t) = Ce in such that y'(t) - Ve or y' - y and thus y(t) is a solution of the differential equation. Show that for any constants C and k, the function y(t) - Ce^is a solution of the differential equation dy/dt = ky. Determine whether y = x^2 is a solution of the differential equation y" - 4xy' + 4y = 0. For what values of r is the function y = e^a solution of the differential equation y" + y' - 6y = 0? For what values of k does the function y = cos(kt) satisfy the differential equation 4y" = -25y? A separable differential equation is a differential equation that can be factored as a function of x times a function of y, for example dy/dt = g(x)h(y). Below is a solution of the separable differential equation dy/dx = x^2/y^2. dy/dx = x^2/y^2 y^2 dy/dx = x^2 y^2 dy = x^2 dx integral y^2 dy = integral x^2 dx y^3/3 + C = x^3/3 + C_1 y^3/3 - x^3/3 = C. Solve the differential equation dy/dx = 6x^2/2y + cos y. Solve the differential equation y' = x^2y. Find the curve that satisfies the differential equation dy/dx + e^(x + y) = 0 and passes through the point (1, 0). The intensity of light, L(x), x feet beneath the surface of the ocean satisfies the differential equation dL/dx = -kL. In the Caribbean Sea diving to a depth of 18 feet cuts the intensity of light in half. At what depth will the intensity of light be one tenth of its surface value?Explanation / Answer
dL/dx = -kL
=> dL / -(kL) = dx
=> x = (-1/k)Log(L)
=> L(x) = e^(-kx)
So, when x = 18, you're told that L is half the intensity.
=> 1/2 = e^(-18k)
=> -Ln(1/2) / 18 = k
k = 0.039
Then you're told that when L falls below one tenth you cannot work without artificial light, and how far down does that occur.
=> 1/10 = e^(-0.039x)
=> -Ln(1/10) / 0.039 = x
x = 59
So, apparently this occurs at about 59 feet.