A-Level Differential equations

Step 1 - differentiate the given substitution, so you know what du/dx looks like. u = xy2, where y is a function of x, so differentiate the product xy2 using the product rule: du/dx = (d/dx of x)(y2) + x(d/dx of y2) du/dx = (1)(y2) + x(d/dx of y2) By the chain rule, d/dx of y2 = 2y (dy/dx), so du/dx = y2 + 2xy (dy/dx) Step 2 - compare that with the equation, which is (2xy dy/dx + y2) cos x = 1/(xy2) The bracket 2xy dy/dx + y2 is exactly du/dx. The denominator on the right, xy2, is exactly u. So the whole equation collapses to (du/dx) cos x = 1/u Step 3 - separate the variables, u on one side and x on the other. Divide both sides by cos x: du/dx = 1/(u cos x) Multiply both sides by u dx: u du = (1/cos x) dx = sec x dx Step 4 - integrate both sides. INT u du = INT sec x dx The left side is a standard power: INT u du = (1/2)u2. The right side is the standard result INT sec x dx = ln|sec x + tan x|. (1/2)u2 = ln|sec x + tan x| + C Step 5 - substitute u = xy2 back and tidy up. (1/2)(xy2)2 = ln|sec x + tan x| + C (1/2)x2y4 = ln|sec x + tan x| + C Multiply through by 2. Since 2C is still just an arbitrary constant, call it C again: x2y4 = 2 ln|sec x + tan x| + C

Method: read the long-term behaviour off the solution of the differential equation found earlier, then draw the curve with that value as a horizontal asymptote. Step 1: The solution obtained earlier has the form C = 18.8 + (C0 - 18.8)e^(-kt), with k > 0, where C0 is the concentration in the tank at t = 0 and 18.8 is the constant (steady-state) term. Step 2: Decide whether the curve rises or falls. C falls, so C0 > 18.8 and the coefficient (C0 - 18.8) is positive. Since k > 0, e^(-kt) decreases from 1 towards 0 as t increases, so C decreases towards 18.8. Step 3: Find the asymptote. As t becomes large, e^(-kt) tends to 0, so C tends to 18.8. The line C = 18.8 is therefore a horizontal asymptote, and C never actually reaches it. Step 4: Sketch, for t >= 0 only, since t is time. - Start the curve on the C-axis at t = 0 at the initial concentration C0, which lies above 18.8. - Draw it decreasing, steeply at first and then more and more gently (concave up). - Draw a dashed horizontal line at C = 18.8, label it, and let the curve flatten onto it from above without crossing it. Step 5: Describe the behaviour in context. For large t the concentration levels off: the rate at which the dissolved substance enters the tank balances the rate at which it leaves, so the tank settles at a steady concentration of about 18.8 mg/L.

The equation is (2xy dy/dx + y²)cos x = 1/(xy²), and the substitution is u = xy². The strategy for a substitution in a differential equation is always the same: differentiate the substitution, then match what you get against the parts of the given equation. Step 1: differentiate u = xy² with respect to x. Here y is a function of x, so xy² is a product of x and y², and both factors depend on x. Use the product rule: du/dx = (d/dx)(x) · y² + x · (d/dx)(y²) and by the chain rule (d/dx)(y²) = 2y dy/dx, so du/dx = y² + x(2y dy/dx) du/dx = y² + 2xy dy/dx. Step 2: match this against the left-hand side of the given equation. The bracket in the equation is (2xy dy/dx + y²), which is exactly du/dx written in the other order. So the left-hand side is cos x · du/dx. Step 3: rewrite the right-hand side in terms of u. The right-hand side is 1/(xy²), and xy² is precisely u, so it is 1/u. Step 4: put the two sides together and rearrange. cos x · du/dx = 1/u Divide both sides by cos x. On 0<x<π/2 we have cos x > 0, so this is allowed and no sign issue arises: du/dx = 1/(u cos x) and since 1/cos x = sec x, du/dx = sec x/u, as required.