A-Level Differential equations: worked solution

Question

Starting from dP/dh=-kP/(293-6.5h) (k>0, P>0), solve to show P=A|293-6.5h|^b for constants A, b, and state whether b is positive or negative with reasons.

Step 1: separate the variables, keeping the minus sign on the right-hand side. dP/dh = -kP/(293 - 6.5h) Divide both sides by P and multiply by dh: (1/P) dP = -k/(293 - 6.5h) dh Integrate both sides: ∫ (1/P) dP = -k ∫ 1/(293 - 6.5h) dh Step 2: do each integral. On the left, P > 0, so ∫ (1/P) dP = ln P (no modulus needed). On the right, the derivative of (293 - 6.5h) is -6.5, so ∫ 1/(293 - 6.5h) dh = -(1/6.5) ln|293 - 6.5h| Multiplying by the -k outside: -k × [-(1/6.5) ln|293 - 6.5h|] = +(k/6.5) ln|293 - 6.5h| So ln P = (k/6.5) ln|293 - 6.5h| + C, where C is the constant of integration. Step 3: undo the logarithm. Use n ln A = ln(A^n) on the first term: ln P = ln(|293 - 6.5h|^(k/6.5)) + C Exponentiate both sides: P = e^C × |293 - 6.5h|^(k/6.5) This is exactly the required form P = A|293 - 6.5h|^b with A = e^C and b = k/6.5 Step 4: state the sign of b, with reasons. k > 0 is given, and 6.5 > 0, so b = k/6.5 is POSITIVE. This also matches the physics. As altitude h increases, the base 293 - 6.5h decreases. Raising a decreasing positive base to a positive power gives a decreasing P, i.e. pressure falls with altitude, which is what the original equation says: for P > 0 and 293 - 6.5h > 0, the right-hand side -kP/(293 - 6.5h) is negative, so dP/dh < 0. A negative b would instead make P increase with altitude, which is impossible.