Questions on first order equations and integrating factors

Question 1

For an equation in the form dy/dx + P(x)y = Q(x), write down the integrating factor and state what multiplying by it achieves.

Accepted Answer

IF = e^∫P dx. After multiplying through, the left side is exactly the derivative of (IF × y), so one integration solves the equation. B1 for the integrating factor, B1 for the product-rule statement.

Question 2

Write down the integrating factor for dy/dx + y tan x = cos x, simplifying it as far as possible.

Accepted Answer

∫tan x dx = ln(sec x), so IF = e^ln sec x = sec x. M1 for ∫tan x dx = ln(sec x), A1 for sec x. Simplify e^ln sec x to sec x; leaving it unsimplified makes every later line harder to handle.

Question 3

Rearrange x dy/dx − 3y = x² (for x > 0) into the standard linear form dy/dx + P(x)y = Q(x), and state P(x) and Q(x).

Accepted Answer

Divide through by x: dy/dx − 3y/x = x, that is dy/dx + P(x)y = Q(x) with P(x) = −3/x and Q(x) = x. M1 for dividing every term by x, A1 for P(x), A1 for Q(x). The coefficient of dy/dx has to be 1 before P and Q can be read off; dividing by anything else leaves a disguised version of the same equation.

Question 4

The general solution of dy/dx + y/x = 2 (for x > 0) is y = x + c/x. Given that y = 4 when x = 1, find c, and verify your value by substituting back into the differential equation.

Accepted Answer

At x = 1: 4 = 1 + c, so c = 3. M1 for substituting the boundary condition, A1 for c = 3. With y = x + 3/x, dy/dx = 1 − 3/x², and y/x = 1 + 3/x². Adding gives dy/dx + y/x = (1 − 3/x²) + (1 + 3/x²) = 2, as required. B1 for the check.

Question 5

Solve dy/dx + y/x = 6x for x > 0, given that y = 3 when x = 1.

Accepted Answer

Here P = 1/x, so IF = e^ln x = x. Multiplying through, d(xy)/dx = 6x². Integrating gives xy = 2x³ + c. The constant belongs on this line, before dividing by x; introducing it afterwards produces cx in place of c. Then y = 3 at x = 1 gives 3 = 2 + c, so c = 1 and y = 2x² + 1/x. Check: dy/dx + y/x = (4x − 1/x²) + (2x + 1/x²) = 6x. M1 A1 for the integrating factor, M1 for xy = 2x³ + c, A1 for c = 1, A1 for the final solution.

Question 6

Find the general solution of dy/dx − 2y = e^5x.

Accepted Answer

P = −2, so IF = e^−2x and d(e^−2xy)/dx = e^3x. The sign of P travels into the exponent unchanged; e^2x here would leave a left side that is nobody's derivative.
Integrating: e^−2xy = e^3x/3 + C, so y = e^5x/3 + Ce^2x. M1 A1 for the integrating factor, M1 for integrating, A1 for the general solution.

Question 7

Solve dy/dx + 2xy = 4x, and describe the behaviour of every solution as x grows.

Accepted Answer

P = 2x is not constant, which changes nothing about the method and only the shape of the factor. IF = e^x², so d(e^x²y)/dx = 4x e^x².
The right side integrates by recognition, since 4x e^x² is the derivative of 2e^x². Hence e^x²y = 2e^x² + C and y = 2 + Ce^−x².
As x grows the transient Ce^−x² dies away very quickly, so every solution settles onto the steady state y = 2, whatever value C takes. M1 for the integrating factor, A1 for e^x², A1 for the general solution, B1 for the limit.

Question 8

Solve dy/dx + 2y = e^−2x, given that y = 3 when x = 0.

Accepted Answer

Here P = 2, so IF = e^2x and multiplying through gives d(e^2xy)/dx = e^2xe^−2x = 1.
Integrating, e^2xy = x + c, so y = (x + c)e^−2x.
Then y = 3 at x = 0 gives c = 3, so y = (x + 3)e^−2x.
M1 for the integrating factor e^2x, A1 for reaching d(e^2xy)/dx = 1, A1 for y = (x + c)e^−2x, M1 for using y = 3 at x = 0, A1 for the particular solution.
Expecting an exponential answer and writing e^2xy = e⁰ + c, or treating the integral of 1 as 0, is what goes wrong.

Question 9

Solve dy/dx + y/(x+1) = 3(x+1) for x > −1, given that y = 5 when x = 0.

Accepted Answer

Here P = 1/(x+1), so IF = e^∫1/(x+1) dx = e^ln(x+1) = x + 1. Multiplying through, d[(x+1)y]/dx = 3(x+1)². M1 A1 for the integrating factor. Integrating, (x+1)y = (x+1)³ + c, so y = (x+1)² + c/(x+1). M1 for integrating, A1 for the general solution. Then y = 5 at x = 0 gives 1 + c = 5, so c = 4, and y = (x+1)² + 4/(x+1). A1 for the particular solution.

Question 10

Solve dy/dx + 4y = 20, given that y = 3 when x = 0. Find, to 3 significant figures, the value of x at which y = 4.9, and explain why y is greater than 4.9 for every larger value of x.

Accepted Answer

P = 4 is constant, so IF = e^4x and d(e^4xy)/dx = 20e^4x. Integrating, e^4xy = 5e^4x + c, so y = 5 + ce^−4x. M1 for the integrating factor, A1 for the general solution. y = 3 at x = 0 gives 5 + c = 3, so c = −2 and y = 5 − 2e^−4x. A1 for the particular solution. Setting y = 4.9: 2e^−4x = 0.1, so e^−4x = 0.05 and x = −(ln 0.05)/4 = 0.749. M1 for the equation in x, A1 for the value. Since dy/dx = 8e^−4x > 0 for every x, y increases throughout, so once it reaches 4.9 it never falls back: every larger x gives a larger y, and so a value above 4.9. B1 for the explanation.

Question 11

A student solving dy/dx − y/x = x for x > 0 writes P = −1/x, then simplifies IF = e^−∫(1/x) dx = e^−ln x to x instead of 1/x, and multiplies the equation through by x. Explain the simplification error, give the correct integrating factor, and show that multiplying by it turns the left-hand side into d(y/x)/dx.

Accepted Answer

e^−ln x = e^ln(1/x) = 1/x, not x: negating a logarithm inverts the quantity inside it, which is a different operation from negating the exponential afterwards. B1 for the error and the correct value. Multiplying dy/dx − y/x = x by 1/x gives (1/x) dy/dx − y/x² = 1. M1 for the correct multiplication. By the quotient rule, d(y/x)/dx = (1/x) dy/dx − y/x², exactly the left-hand side reached, so the equation becomes d(y/x)/dx = 1. M1 A1 for identifying the derivative and the resulting equation. Multiplying by x instead would leave x dy/dx − y = x², whose left side is not the derivative of any single product, and the method stalls.

Question 12

Solve dy/dx − 2y/x = x² ln x for x > 0, given that y = 0 when x = 1.

Accepted Answer

P = −2/x, so ∫P dx = −2 ln x and IF = e^{−2 ln x} = x⁻². Reading the factor as e^{−2 ln x} without simplifying to a power of x is what makes the next integration look impossible.
Multiplying through, d(x⁻²y)/dx = ln x.
Now integrate ln x by parts, taking u = ln x and dv = dx: ∫ln x dx = x ln x − ∫1 dx = x ln x − x + c. Writing ln x as 1 × ln x is the step worth spotting; there is no integral of ln x by inspection.
So x⁻²y = x ln x − x + c, and y = 0 at x = 1 gives 0 = 0 − 1 + c, so c = 1.
Hence y = x²(x ln x − x + 1), that is x³ ln x − x³ + x².
Check at x = 2: y = 8 ln 2 − 8 + 4 ≈ 1.545, and differentiating the answer returns the original equation.
M1 A1 for the integrating factor, M1 for reaching d(x⁻²y)/dx = ln x, M1 A1 for the parts integration, A1 for the particular solution.

Question 13

A tank holds 100 litres of water in which 5 kg of salt is dissolved. Brine containing 0.2 kg of salt per litre runs in at 4 litres per minute, and the mixture, kept uniform by stirring, runs out at 4 litres per minute. Let S kg be the mass of salt in the tank after t minutes. Show that dS/dt + 0.04S = 0.8, solve the equation, and find the time at which the tank holds 15 kg of salt, giving your answer to 3 significant figures.

Accepted Answer

Salt enters at 0.2 × 4 = 0.8 kg per minute. The two flow rates are equal, so the volume stays at 100 litres and the concentration leaving is S/100 kg per litre, carrying salt out at 4 × S/100 = 0.04S kg per minute.
Hence dS/dt = 0.8 − 0.04S, that is dS/dt + 0.04S = 0.8 as required.
The integrating factor is e^0.04t, so d(e^0.04tS)/dt = 0.8e^0.04t and e^0.04tS = 20e^0.04t + A. Therefore S = 20 + Ae^−0.04t.
At t = 0, S = 5 gives A = −15, so S = 20 − 15e^−0.04t.
Setting S = 15: 15e^−0.04t = 5, so e^−0.04t = 1/3 and 0.04t = ln 3. Then t = 25 ln 3 = 27.5 minutes to 3 significant figures.
M1 for rate in minus rate out, A1 for dS/dt = 0.8 − 0.04S, M1 for the integrating factor e^0.04t, A1 for S = 20 + Ae^−0.04t, A1 for A = −15, M1 for solving 15 = 20 − 15e^−0.04t, A1 for 27.5 minutes.
The long-run mass is 20 kg, which is 0.2 × 100: the tank eventually holds brine at the incoming concentration. The model assumes stirring makes the mixture uniform the instant salt enters, which no real tank achieves, and it relies on the inflow and outflow rates being equal, since unequal rates would make the volume, and so the coefficient of S, a function of t.

Question 14

Solve dy/dx = 1/(x + y), given that y = 0 when x = 0. Give your answer in the form x = f(y).

Accepted Answer

As it stands this is neither separable nor linear in y, since y appears inside the reciprocal. Taking reciprocals turns it round: dx/dy = x + y, which is linear in x, with y now the independent variable.
Write it as dx/dy − x = y. Here P = −1, so IF = e^−y and d(e^−yx)/dy = ye^−y.
Integrate the right side by parts, with u = y and dv = e^−ydy: ∫ye^−y dy = −ye^−y + ∫e^−y dy = −ye^−y − e^−y + c = −(y + 1)e^−y + c.
So e^−yx = −(y + 1)e^−y + c, and multiplying through by e^y gives x = −(y + 1) + ce^y.
The condition x = 0 at y = 0 gives 0 = −1 + c, so c = 1 and x = e^y − y − 1.
M1 for taking reciprocals to obtain dx/dy = x + y, A1 for dx/dy − x = y, M1 for the integrating factor e^−y, A1 for the parts integration, M1 for applying the condition, A1 for x = e^y − y − 1.
Checking: dx/dy = e^y − 1, while x + y = e^y − 1 as well. An answer left as e^−yx = −(y + 1)e^−y + c has not been rearranged into the requested form.

Question 15

Solve dy/dx + y = x², given that y = 1 when x = 0.

Accepted Answer

P = 1, so IF = e^x and d(e^xy)/dx = x²e^x. M1 A1 for the integrating factor. Integrating x²e^x needs integration by parts twice: taking u = x², dv = e^xdx gives ∫x²e^x dx = x²e^x − ∫2xe^x dx, and taking u = 2x, dv = e^xdx on the remaining integral gives ∫2xe^x dx = 2xe^x − 2e^x + c. So ∫x²e^x dx = e^x(x² − 2x + 2) + c. M1 A1 for the repeated integration by parts. Hence e^xy = e^x(x² − 2x + 2) + c, so y = x² − 2x + 2 + ce^−x. Then y = 1 at x = 0 gives 2 + c = 1, so c = −1, and y = x² − 2x + 2 − e^−x. M1 for applying the condition, A1 for the particular solution. Each pass of parts removes one power of x from the polynomial factor; stopping after one pass here would leave an integral, xe^x, still unresolved.