ion $C$ gives:\n\n$$4\\ln|x + 2| + \\ln|x - 1| + C$$\n\n*(This earns the accuracy mark **A1**)*"}]}],"generatedAt":"2026-06-09T06:13:48.987Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null},{"id":"741a759c-3ead-4636-ad9e-d07c6a15f154","revisionId":"d53da95d-829f-4530-b214-02db368b0704","questionNumber":null,"questionType":"LA","specification":"","difficulty":2,"difficultyBand":"medium","totalMarks":6,"sourceTitle":null,"calculatorUse":null,"parts":[{"id":"4f3cd506-6d57-49a3-9926-4b0ae3945c95","label":"a","content":"Find the solution to the differential equation $\\frac{dy}{dx} = (y - 3)^2$ given that when $y = 4$, $x = 2$\n\nGive your answer in the form $y = f(x)$","markscheme":"- Separates variables: $\\int (y-3)^{-2} dy = \\int 1 dx$ M1\n- Integrates: $-(y-3)^{-1} = x + c$ A2\n- Substitutes $y=4, x=2$: $-(1)^{-1} = 2 + c \\implies c = -3$ M1\n- Rearranges: $-\\frac{1}{y-3} = x - 3 \\implies y-3 = -\\frac{1}{x-3}$ M1\n- Final answer: $y = 3 - \\frac{1}{x-3}$ or $y = \\frac{3x-10}{x-3}$ A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Evidence of separating variables, typically $\\int (y-3)^{-2} dy = \\int dx$.","code":"M"},{"id":"p2","marks":1,"description":"Correct integration of the LHS to get $-(y-3)^{-1}$ or equivalent.","code":"A"},{"id":"p3","marks":1,"description":"Correct integration of the RHS to get $x + c$ or $x$ with a constant included later.","code":"A"},{"id":"p4","marks":1,"description":"Substitution of $y=4$ and $x=2$ into an integrated expression to find the value of $c$.","code":"M"},{"id":"p5","marks":1,"description":"Attempt to rearrange the equation to make $y$ the subject.","code":"M"},{"id":"p6","marks":1,"description":"Final correct expression in the form $y = f(x)$.","code":"A"}],"finalAnswers":["y = 3 - \\frac{1}{x-3}","y = \\frac{3x-10}{x-3}","y = \\frac{10-3x}{3-x}"],"generatedAt":"2026-05-06T14:04:04.203Z","modelVersion":"gemini-3-flash-preview"},"explanation":{"methods":[{"id":"indefinite_integration","label":"Indefinite Integration","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve this first-order differential equation, we need to separate the variables so that all terms involving $y$ are on one side with $dy$, and all terms involving $x$ are on the other side with $dx$.\n\nGiven equation:\n$$\\frac{dy}{dx} = (y - 3)^2$$\n\nDividing both sides by $(y - 3)^2$ and multiplying by $dx$ gives:\n$$\\frac{1}{(y - 3)^2} \\, dy = 1 \\, dx$$\n\nTo prepare this for integration, we rewrite the fraction using a negative index:\n$$(y - 3)^{-2} \\, dy = 1 \\, dx$$\n\nThis earns the first method mark (**M1**)."},{"type":"text","order":1,"title":"Integrate both sides","content":"Now we integrate both sides of our separated equation:\n$$\\int (y - 3)^{-2} \\, dy = \\int 1 \\, dx$$\n\nWe integrate the left-hand side using the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ (since the derivative of $y-3$ with respect to $y$ is just $1$):\n$$\\frac{(y-3)^{-1}}{-1} = x + c$$\n\nSimplifying this yields:\n$$-(y - 3)^{-1} = x + c$$\n\nOr in fractional form:\n$$-\\frac{1}{y - 3} = x + c$$\n\nThis yields two accuracy marks (**A2**) for obtaining the correct integrated general expression."},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$-\\frac{1}{\\htmlData{anchor=y-sub}{\\color{@sky}{y}} - 3} = \\htmlData{anchor=x-sub}{\\color{@amber}{x}} + c$"},{"latex":"$$-\\frac{1}{\\htmlData{anchor=y-val}{\\color{@sky}{4}} - 3} = \\htmlData{anchor=x-val}{\\color{@amber}{2}} + c$"},{"latex":"$$-\\frac{1}{1} = 2 + c \\implies -1 = 2 + c$"},{"latex":"$$c = \\htmlData{anchor=c-val}{\\color{@purple}{-3}}$"}],"highlights":[{"color":"sky","style":"box","anchor":"y-val"},{"color":"amber","style":"box","anchor":"x-val"},{"color":"purple","style":"text-color","anchor":"c-val"}],"connections":[{"to":"y-sub","from":"y-val","color":"sky","label":"y = 4","style":"solid"},{"to":"x-sub","from":"x-val","color":"

Integration

Question 23

Integration Questions

Integration

Question 23

Integration Questions