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5 + c \\implies c = -6$ M1\n- Rearranges: $-\\frac{1}{y-1} = x - 6 \\implies y-1 = \\frac{1}{6-x}$ M1\n- Final answer: $y = 1 + \\frac{1}{6-x}$ or $y = \\frac{7-x}{6-x}$ A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Separation of variables seen.","code":"M"},{"id":"p2","marks":1,"description":"Integrates LHS to $-(y-1)^{-1}$.","code":"A"},{"id":"p3","marks":1,"description":"Integrates RHS to $x + c$.","code":"A"},{"id":"p4","marks":1,"description":"Correct substitution of boundary conditions $(5, 2)$ to find $c$.","code":"M"},{"id":"p5","marks":1,"description":"Rearranges the resulting equation to make $y$ the subject.","code":"M"},{"id":"p6","marks":1,"description":"Final answer in the required form $y = f(x)$.","code":"A"}],"finalAnswers":["y = 1 + \\frac{1}{6-x}","y = \\frac{7-x}{6-x}"],"generatedAt":"2026-05-06T14:04:07.869Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"indefinite_integration","label":"Indefinite Integration","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation \n\n$$\\frac{dy}{dx} = (y - 1)^2$$\n\nwe first separate the variables. We bring all terms involving $y$ to the left side with $dy$, and all terms involving $x$ to the right side with $dx$:\n\n$$\\frac{1}{(y - 1)^2} \\, dy = 1 \\, dx$$\n\nThis can be written using a negative index as:\n\n$$(y - 1)^{-2} \\, dy = dx$$"},{"type":"text","order":1,"title":"Integrate both sides","content":"Now we integrate both sides of the equation:\n\n$$\\int (y - 1)^{-2} \\, dy = \\int 1 \\, dx$$\n\nUsing the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ for $n \\neq -1$, we get:\n\n$$-\\frac{1}{y - 1} = x + c$$\n\nwhere $c$ is the arbitrary constant of integration."},{"type":"text","order":2,"title":"Find the constant $c$","content":"We are given the initial conditions that $y = 2$ when $x = 5$. We substitute these values into our integrated equation to solve for $c$:\n\n$$-\\frac{1}{2 - 1} = 5 + c$$\n\n$$-1 = 5 + c \\implies c = -6$$"},{"type":"text","order":3,"title":"Substitute $c$ and rearrange","content":"Substituting $c = -6$ back into the integrated equation gives:\n\n$$-\\frac{1}{y - 1} = x - 6$$\n\nNow we rearrange this equation to express $y$ in terms of $x$. First, multiply both sides by $-1$ to make the fraction positive:\n\n$$\\frac{1}{y - 1} = 6 - x$$\n\nNext, take the reciprocal of both sides:\n\n$$y - 1 = \\frac{1}{6 - x}$$\n\nFinally, add $1$ to both sides to solve for $y$:\n\n$$y = 1 + \\frac{1}{6 - x}$$"},{"type":"text","order":4,"title":"Write the final answer","content":"We can express our final answer in either of the following equivalent forms:\n\n$$y = 1 + \\frac{1}{6-x}$$\n\nOr, by finding a common denominator:\n\n$$y = \\frac{6-x}{6-x} + \\frac{1}{6-x} = \\frac{7-x}{6-x}$$"}]},{"id":"definite_integration","label":"Definite Integration with Limits","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation \n\n$$\\frac{dy}{dx} = (y - 1)^2$$\n\nwe first separate the variables so that all terms involving $y$ are on one side and all terms involving $x$ are on the other side. \n\nDividing both sides by $(y - 1)^2$ and multiplying by $dx$ gives:\n\n$$\\frac{1}{(y - 1)^2} \\, dy = 1 \\, dx$$"},{"type":"text","order":1,"title":"Set up definite integrals","content":"Using the **Definite Integration with Limits** approach, we integrate both sides. \n\nWe set the lower limits of integration to the given initial condition values, $y = 2$ and $x = 5$, and the upper limits to the general variables, $y$ and $x$. \n\nTo avoid using the same letter for the integration variable and the limit, we can use dummy variables $u$ and $v$:\n\n$$\\int_{2}^{y} \\frac{1}{(u - 1)^2} \\, du = \\int_{5}^{x} 1 \\, dv$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{LHS: } \\int_{2}^{y} (u - 1)^{-2} \\, du = \\left[ \\htmlData{anchor=int-u}{\\color{@sky}{-\\frac{1}{u - 1}}} \\right]_{2}^{y} = \\htmlData{anchor=sub-y}{\\color{@sky} {-\\frac{1}{y - 1}}} - \\left( \\htmlData{anchor=sub-2}{\\color{@pink}{-\\frac{1}{2 - 1}}} \\right)$"},{"latex":"$$\\text{RHS: } \\int_{5}^{x} 1 \\, dv = \\left[ v \\right]_{5}^{x} = x - 5$"}],"highlights":[{"color":"sky","style":"box","anchor":"int-u"},{"color":"sky","style":"text-color","anchor":"sub-y"},{"color":"pink","style":"text-color","anchor":"sub-2"}],"connections":[{"to":"sub-y","from":"int-u","color":"sky","label":"substitute y","style":"solid"},{"to":"sub-2","from":"int-u","color":"pink","label":"substitute 2","style":"dashed"}]},"order":2,"title":"Integrate both sides","description":"We perform the integration on both sides using standard integration rules."},{"type":"text","order":3,"title":"Evaluate the limits","content":"Let's simplify our integrated expressions:\n\n$$\\text{LHS} = -\\frac{1}{y - 1} - (-1) = -\\frac{1}{y - 1} + 1$$\n\n$$\\text{RHS} = x - 5$$\n\nEquating both sides, we get:\n\n$$-\\frac{1}{y - 1} + 1 = x - 5$$"},{"type":"text","order":4,"title":"Rearrange to solve for y","content":"Now, we rearrange the equation directly to make $y$ the subject. Subtract $1$ from both sides:\n\n$$-\\frac{1}{y - 1} = x - 6$$\n\nMultiply by $-1$:\n\n$$\\frac{1}{y - 1} = 6 - x$$\n\nTaking the reciprocal of both sides:\n\n$$y - 1 = \\frac{1}{6 - x}$$"},{"type":"text","order":5,"title":"Write the final answer","content":"Finally, add $1$ to both sides to solve for $y$:\n\n$$y = 1 + \\frac{1}{6 - x}$$\n\nAlternatively, we can write this under a single common denominator:\n\n$$y = \\frac{6 - x + 1}{6 - x} = \\frac{7 - x}{6 - x}$$\n\nBoth forms are completely correct."}]}],"generatedAt":"2026-06-09T04:45:19.481Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"c8a40776-d1a6-4000-970d-d3f8f0e43bb1:indefinite_integration","label":"Indefinite Integration","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation \n\n$$\\frac{dy}{dx} = (y - 1)^2$$\n\nwe first separate the variables. We bring all terms involving $y$ to the left side with $dy$, and all terms involving $x$ to the right side with $dx$:\n\n$$\\frac{1}{(y - 1)^2} \\, dy = 1 \\, dx$$\n\nThis can be written using a negative index as:\n\n$$(y - 1)^{-2} \\, dy = dx$$"},{"type":"text","order":1,"title":"Integrate both sides","content":"Now we integrate both sides of the equation:\n\n$$\\int (y - 1)^{-2} \\, dy = \\int 1 \\, dx$$\n\nUsing the power rule for integration, $\\int u^n \\, du = \\frac{u^{n+1}}{n+1}$ for $n \\neq -1$, we get:\n\n$$-\\frac{1}{y - 1} = x + c$$\n\nwhere $c$ is the arbitrary constant of integration."},{"type":"text","order":2,"title":"Find the constant $c$","content":"We are given the initial conditions that $y = 2$ when $x = 5$. We substitute these values into our integrated equation to solve for $c$:\n\n$$-\\frac{1}{2 - 1} = 5 + c$$\n\n$$-1 = 5 + c \\implies c = -6$$"},{"type":"text","order":3,"title":"Substitute $c$ and rearrange","content":"Substituting $c = -6$ back into the integrated equation gives:\n\n$$-\\frac{1}{y - 1} = x - 6$$\n\nNow we rearrange this equation to express $y$ in terms of $x$. First, multiply both sides by $-1$ to make the fraction positive:\n\n$$\\frac{1}{y - 1} = 6 - x$$\n\nNext, take the reciprocal of both sides:\n\n$$y - 1 = \\frac{1}{6 - x}$$\n\nFinally, add $1$ to both sides to solve for $y$:\n\n$$y = 1 + \\frac{1}{6 - x}$$"},{"type":"text","order":4,"title":"Write the final answer","content":"We can express our final answer in either of the following equivalent forms:\n\n$$y = 1 + \\frac{1}{6-x}$$\n\nOr, by finding a common denominator:\n\n$$y = \\frac{6-x}{6-x} + \\frac{1}{6-x} = \\frac{7-x}{6-x}$$"}]},{"id":"c8a40776-d1a6-4000-970d-d3f8f0e43bb1:definite_integration","label":"Definite Integration with Limits","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation \n\n$$\\frac{dy}{dx} = (y - 1)^2$$\n\nwe first separate the variables so that all terms involving $y$ are on one side and all terms involving $x$ are on the other side. \n\nDividing both sides by $(y - 1)^2$ and multiplying by $dx$ gives:\n\n$$\\frac{1}{(y - 1)^2} \\, dy = 1 \\, dx$$"},{"type":"text","order":1,"title":"Set up definite integrals","content":"Using the **Definite Integration with Limits** approach, we integrate both sides. \n\nWe set the lower limits of integration to the given initial condition values, $y = 2$ and $x = 5$, and the upper limits to the general variables, $y$ and $x$. \n\nTo avoid using the same letter for the integration variable and the limit, we can use dummy variables $u$ and $v$:\n\n$$\\int_{2}^{y} \\frac{1}{(u - 1)^2} \\, du = \\int_{5}^{x} 1 \\, dv$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{LHS: } \\int_{2}^{y} (u - 1)^{-2} \\, du = \\left[ \\htmlData{anchor=int-u}{\\color{@sky}{-\\frac{1}{u - 1}}} \\right]_{2}^{y} = \\htmlData{anchor=sub-y}{\\color{@sky} {-\\frac{1}{y - 1}}} - \\left( \\htmlData{anchor=sub-2}{\\color{@pink}{-\\frac{1}{2 - 1}}} \\right)$"},{"latex":"$$\\text{RHS: } \\int_{5}^{x} 1 \\, dv = \\left[ v \\right]_{5}^{x} = x - 5$"}],"highlights":[{"color":"sky","style":"box","anchor":"int-u"},{"color":"sky","style":"text-color","anchor":"sub-y"},{"color":"pink","style":"text-color","anchor":"sub-2"}],"connections":[{"to":"sub-y","from":"int-u","color":"sky","label":"substitute y","style":"solid"},{"to":"sub-2","from":"int-u","color":"pink","label":"substitute 2","style":"dashed"}]},"order":2,"title":"Integrate both sides","description":"We perform the integration on both sides using standard integration rules."},{"type":"text","order":3,"title":"Evaluate the limits","content":"Let's simplify our integrated expressions:\n\n$$\\text{LHS} = -\\frac{1}{y - 1} - (-1) = -\\frac{1}{y - 1} + 1$$\n\n$$\\text{RHS} = x - 5$$\n\nEquating both sides, we get:\n\n$$-\\frac{1}{y - 1} + 1 = x - 5$$"},{"type":"text","order":4,"title":"Rearrange to solve for y","content":"Now, we rearrange the equation directly to make $y$ the subject. Subtract $1$ from both sides:\n\n$$-\\frac{1}{y - 1} = x - 6$$\n\nMultiply by $-1$:\n\n$$\\frac{1}{y - 1} = 6 - x$$\n\nTaking the reciprocal of both sides:\n\n$$y - 1 = \\frac{1}{6 - x}$$"},{"type":"text","order":5,"title":"Write the final answer","content":"Finally, add $1$ to both sides to solve for $y$:\n\n$$y = 1 + \\frac{1}{6 - x}$$\n\nAlternatively, we can write this under a single common denominator:\n\n$$y = \\frac{6 - x + 1}{6 - x} = \\frac{7 - x}{6 - x}$$\n\nBoth forms are completely correct."}]}],"generatedAt":"2026-06-09T04:45:29.558Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"7ac60a40-5875-41d5-aac8-895b765c47d4","revisionId":"257f21eb-12b0-4dfc-8479-e55deca42a20","questionNumber":null,"questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"f3ed6a61-7e58-414a-aeba-675e720a9e19","label":"a","content":"Find the general solution to the differential equation $$\\frac{dy}{dx} = e^y \\ln x$$","markscheme":"- Separates variables: $\\int e^{-y} dy = \\int \\ln x dx$ M1\n- Integrates LHS: $-e^{-y}$ A1\n- Integrates RHS by parts: $u=\\ln x, v'=1 \\implies u'=1/x, v=x$ M1\n- RHS $= x \\ln x - \\int 1 dx = x \\ln x - x + c$ A2\n- Final solution: $-e^{-y} = x \\ln x - x + c$ A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Evidence of separating variables leading to $\\int e^{-y} dy$.","code":"M"},{"id":"p2","marks":1,"description":"Correct integration of the LHS to $-e^{-y}$.","code":"A"},{"id":"p3","marks":1,"description":"Correct setup for RHS integration by parts.","code":"M"},{"id":"p4","marks":2,"description":"Correct integration of the RHS to $x \\ln x - x + c$.","code":"A"},{"id":"p5","marks":1,"description":"Final equation correctly relating the integrated LHS and RHS.","code":"A"}],"finalAnswers":["-e^{-y} = x \\ln x - x + c","e^{-y} = x - x \\ln x + K"],"generatedAt":"2026-05-06T14:04:20.674Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"separation_of_variables","label":"Separation of Variables","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation:\n\n$$\\frac{dy}{dx} = e^y \\ln x$$\n\nwe use the method of **separation of variables**. We group all terms involving $y$ on the left-hand side (LHS) and all terms involving $x$ on the right-hand side (RHS):\n\n$$\\frac{1}{e^y} dy = \\ln x \\, dx$$\n\nRewriting $\\frac{1}{e^y}$ as $e^{-y}$, we set up the integrals on both sides:\n\n$$\\int e^{-y} \\, dy = \\int \\ln x \\, dx$$"},{"type":"text","order":1,"title":"Integrate the LHS","content":"Now we integrate the left-hand side directly using standard integration rules:\n\n$$\\int e^{-y} \\, dy = -e^{-y}$$"},{"type":"text","order":2,"title":"Integrate the RHS","content":"To integrate the right-hand side, we use **integration by parts**:\n\n$$\\int u \\, dv = uv - \\int v \\, du$$\n\nLet us choose:\n- $u = \\ln x \\implies du = \\frac{1}{x} \\, dx$\n- $dv = dx \\implies v = x$\n\nApplying the integration by parts formula:\n\n$$\\int \\ln x \\, dx = x \\ln x - \\int x \\left(\\frac{1}{x}\\right) dx$$\n\nSimplifying the integral on the right:\n\n$$\\int \\ln x \\, dx = x \\ln x - \\int 1 \\, dx$$\n\n$$\\int \\ln x \\, dx = x \\ln x - x$$"},{"type":"text","order":3,"title":"State the general solution","content":"Equating the two integrated sides and adding the constant of integration $c$, we obtain the general solution:\n\n$$-e^{-y} = x \\ln x - x + c$$"}]}],"generatedAt":"2026-06-09T06:14:05.247Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"f3ed6a61-7e58-414a-aeba-675e720a9e19:separation_of_variables","label":"Separation of Variables","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation:\n\n$$\\frac{dy}{dx} = e^y \\ln x$$\n\nwe use the method of **separation of variables**. We group all terms involving $y$ on the left-hand side (LHS) and all terms involving $x$ on the right-hand side (RHS):\n\n$$\\frac{1}{e^y} dy = \\ln x \\, dx$$\n\nRewriting $\\frac{1}{e^y}$ as $e^{-y}$, we set up the integrals on both sides:\n\n$$\\int e^{-y} \\, dy = \\int \\ln x \\, dx$$"},{"type":"text","order":1,"title":"Integrate the LHS","content":"Now we integrate the left-hand side directly using standard integration rules:\n\n$$\\int e^{-y} \\, dy = -e^{-y}$$"},{"type":"text","order":2,"title":"Integrate the RHS","content":"To integrate the right-hand side, we use **integration by parts**:\n\n$$\\int u \\, dv = uv - \\int v \\, du$$\n\nLet us choose:\n- $u = \\ln x \\implies du = \\frac{1}{x} \\, dx$\n- $dv = dx \\implies v = x$\n\nApplying the integration by parts formula:\n\n$$\\int \\ln x \\, dx = x \\ln x - \\int x \\left(\\frac{1}{x}\\right) dx$$\n\nSimplifying the integral on the right:\n\n$$\\int \\ln x \\, dx = x \\ln x - \\int 1 \\, dx$$\n\n$$\\int \\ln x \\, dx = x \\ln x - x$$"},{"type":"text","order":3,"title":"State the general solution","content":"Equating the two integrated sides and adding the constant of integration $c$, we obtain the general solution:\n\n$$-e^{-y} = x \\ln x - x + c$$"}]}],"generatedAt":"2026-06-09T06:14:13.213Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"8a13ecc5-a8a9-4fa9-8108-320bd7416951","revisionId":"5cbb9a97-b39b-4d87-b906-ba3d678090c4","questionNumber":"4","questionType":"LA","specification":"","difficultyBand":"medium","tier":null,"totalMarks":6,"sourceTitle":"11.10 Solving Differential Equations","calculatorUse":null,"isCalculation":true,"parts":[{"id":"411a67dc-b528-42da-9389-6bbcc097daaa","label":"0","content":"Find the general solution to the differential equation $\\frac{dy}{dx} = xy \\sin x$","markscheme":"- Separates variables for \$y \\ne 0\$: \$\\int \\frac{1}{y} dy = \\int x \\sin x dx\$ M1\n- Integrates LHS: \$\\ln|y|\$ B1\n- Integrates RHS by parts: \$u=x, v'=\\sin x \\implies u'=1, v=-\\cos x\$ M1\n- RHS \$= -x \\cos x - \\int -\\cos x dx = -x \\cos x + \\sin x + c\$ A2\n- Final solution: \$y = C e^{\\sin x - x \\cos x}\$, where \$C\$ is any real constant, including \$C=0\$ A1","marks":6,"order":0,"markingRubric":{"totalMarks":6,"points":[{"id":"p1","marks":1,"description":"Separates variables correctly.","code":"M"},{"id":"p2","marks":1,"description":"Integrates $\\frac{1}{y}$ to $\\ln y$.","code":"B"},{"id":"p3","marks":1,"description":"Correctly identifies $u$ and $dv/dx$ for integration by parts.","code":"M"},{"id":"p4","marks":2,"description":"Correctly integrates $x \\sin x$ to $-x \\cos x + \\sin x$.","code":"A"},{"id":"p5","marks":1,"description":"Includes the constant of integration.","code":"A"}],"finalAnswers":["\\ln y = -x \\cos x + \\sin x + c"],"generatedAt":"2026-05-06T03:36:05.789Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"separation_of_variables","label":"Separation of Variables","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation using the method of **Separation of Variables**, we group all terms involving $y$ on the left-hand side (LHS) and all terms involving $x$ on the right-hand side (RHS):\n\n$$\\frac{dy}{dx} = xy \\sin x$$\n\nDividing both sides by $y$ and multiplying both sides by $dx$ gives:\n\n$$\\frac{1}{y} \\, dy = x \\sin x \\, dx$$\n\nNow, we write this in integral form:\n\n$$\\int \\frac{1}{y} \\, dy = \\int x \\sin x \\, dx$$"},{"type":"text","order":1,"title":"Integrate the left-hand side","content":"The integral on the left-hand side is a standard integral:\n\n$$\\int \\frac{1}{y} \\, dy = \\ln |y|$$\n\nFor general solutions, we can write this simply as $\\ln y$."},{"type":"text","order":2,"title":"Set up integration by parts","content":"To integrate the right-hand side, $\\int x \\sin x \\, dx$, we must use **Integration by Parts**. \n\nThe integration by parts formula is:\n\n$$\\int u \\frac{dv}{dx} \\, dx = uv - \\int v \\frac{du}{dx} \\, dx$$\n\nLet's choose our parts:\n* Let $u = x \\implies \\frac{du}{dx} = 1$\n* Let $\\frac{dv}{dx} = \\sin x \\implies v = -\\cos x$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\htmlData{anchor=u-part}{\\color{@sky}{x}} \\htmlData{anchor=dv-part}{\\color{@amber}{\\sin x}} \\, dx = \\htmlData{anchor=uv-product}{\\color{@sky}{x} (\\color{@amber}{-\\cos x})} - \\int \\htmlData{anchor=vdu-product}{(\\color{@amber}{-\\cos x}) (\\color{@sky}{1})} \\, dx$"},{"latex":"$$\\phantom{\\int x \\sin x \\, dx} = -x \\cos x + \\int \\cos x \\, dx$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-part"},{"color":"amber","style":"text-color","anchor":"dv-part"}],"connections":[{"to":"uv-product","from":"u-part","color":"sky","label":"u","style":"solid"},{"to":"vdu-product","from":"dv-part","color":"amber","label":"v","style":"solid"}]},"order":3,"title":"Apply integration by parts","description":"Substitute our chosen parts $u$, $v$, $\\frac{du}{dx}$, and $\\frac{dv}{dx}$ into the integration by parts formula."},{"type":"text","order":4,"title":"Complete the integration","content":"Now we integrate the remaining term on the right-hand side, remembering to add a constant of integration $c$:\n\n$$\\int \\cos x \\, dx = \\sin x$$\n\nPutting this together, the right-hand side becomes:\n\n$$\\int x \\sin x \\, dx = -x \\cos x + \\sin x + c$$"},{"type":"text","order":5,"title":"State the general solution","content":"By equating our integrated left-hand side and right-hand side, we obtain the general solution:\n\n$$\\ln y = -x \\cos x + \\sin x + c$$"}]}],"generatedAt":"2026-06-09T06:43:28.727Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"411a67dc-b528-42da-9389-6bbcc097daaa:separation_of_variables","label":"Separation of Variables","steps":[{"type":"text","order":0,"title":"Separate the variables","content":"To solve the differential equation using the method of **Separation of Variables**, we group all terms involving \$y\$ on the left-hand side (LHS) and all terms involving \$x\$ on the right-hand side (RHS). For \$y \\ne 0\$, we can divide by \$y\$:\n\n$$\\frac{dy}{dx} = xy \\sin x$$\n\nDividing both sides by \$y\$ and multiplying both sides by \$dx\$ gives:\n\n$$\\frac{1}{y} \\, dy = x \\sin x \\, dx$$\n\nNow, we write this in integral form:\n\n$$\\int \\frac{1}{y} \\, dy = \\int x \\sin x \\, dx$$\n\nThe solution \$y=0\$ will be included separately in the final family of solutions."},{"type":"text","order":1,"title":"Integrate the left-hand side","content":"The integral on the left-hand side is:\n\n$$\\int \\frac{1}{y} \\, dy = \\ln |y|$$\n\nWe must retain the absolute value because the general solution may have \$y<0\$."},{"type":"text","order":2,"title":"Set up integration by parts","content":"To integrate the right-hand side, $\\int x \\sin x \\, dx$, we must use **Integration by Parts**. \n\nThe integration by parts formula is:\n\n$$\\int u \\frac{dv}{dx} \\, dx = uv - \\int v \\frac{du}{dx} \\, dx$$\n\nLet's choose our parts:\n* Let $u = x \\implies \\frac{du}{dx} = 1$\n* Let $\\frac{dv}{dx} = \\sin x \\implies v = -\\cos x$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\int \\htmlData{anchor=u-part}{\\color{@sky}{x}} \\htmlData{anchor=dv-part}{\\color{@amber}{\\sin x}} \\, dx = \\htmlData{anchor=uv-product}{\\color{@sky}{x} (\\color{@amber}{-\\cos x})} - \\int \\htmlData{anchor=vdu-product}{(\\color{@amber}{-\\cos x}) (\\color{@sky}{1})} \\, dx$"},{"latex":"$$\\phantom{\\int x \\sin x \\, dx} = -x \\cos x + \\int \\cos x \\, dx$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"u-part"},{"color":"amber","style":"text-color","anchor":"dv-part"}],"connections":[{"to":"uv-product","from":"u-part","color":"sky","label":"u","style":"solid"},{"to":"vdu-product","from":"dv-part","color":"amber","label":"v","style":"solid"}]},"order":3,"title":"Apply integration by parts","description":"Substitute our chosen parts $u$, $v$, $\\frac{du}{dx}$, and $\\frac{dv}{dx}$ into the integration by parts formula."},{"type":"text","order":4,"title":"Complete the integration","content":"Now we integrate the remaining term on the right-hand side, remembering to add a constant of integration $c$:\n\n$$\\int \\cos x \\, dx = \\sin x$$\n\nPutting this together, the right-hand side becomes:\n\n$$\\int x \\sin x \\, dx = -x \\cos x + \\sin x + c$$"},{"type":"text","order":5,"title":"State the general solution","content":"By equating our integrated left-hand side and right-hand side, we obtain:\n\n$$\\ln |y| = -x \\cos x + \\sin x + c$$\n\nExponentiating both sides gives:\n\n$$|y| = e^c e^{\\sin x - x\\cos x}$$\n\nThe sign can be absorbed into an arbitrary real constant \$C\$, and \$C=0\$ gives the separately valid solution \$y=0\$. Therefore, the general solution is:\n\n$$y = C e^{\\sin x - x \\cos x}, \\qquad C \\in \\mathbb{R}$$"}]}],"generatedAt":"2026-06-09T06:43:38.334Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null}],"subjectCode":"a_level_maths","summaries":[{"id":"3f24699d-1671-4687-91a3-3aa389d44093","difficultyBand":"hard","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"4c1f1bce-ef43-47fa-b033-0e73d2b05a35","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"741a759c-3ead-4636-ad9e-d07c6a15f154","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"262b28a5-4173-4856-91f4-af14688fb643","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"c0f605e7-1d0a-43f0-8307-bdc4e0657a19","difficultyBand":"hard","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"d2d74518-ad96-4bb8-931a-2b1e293a3b43","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"e8b31581-1c86-46f5-93c9-3848c44cfc14","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"11142360-15b8-4cca-9567-969d38d77fbd","difficultyBand":"hard","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"21e6151d-5f2d-4f22-a72e-97bb1d271f49","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"3625a5ec-228a-4dbb-a8ee-3fa9ee4db663","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"613a5699-3205-4cdb-866f-0a4886c7711c","difficultyBand":"hard","calculatorUse":null,"totalMarks":12,"questionType":"LA","isCalculation":true},{"id":"7ac60a40-5875-41d5-aac8-895b765c47d4","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"8a13ecc5-a8a9-4fa9-8108-320bd7416951","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"f6764fc3-10cd-4a99-84b4-3b35389878f9","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"5b129fd6-a6bc-43ef-9b14-7a303cc7e99a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"02c8c314-1f58-4bdb-bc0c-95bd49165f34","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"0f965470-38b2-4afc-b53d-4b22ce238170","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"1cd59b43-aecb-405e-a64a-f0e1e4e67b79","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"1df5c3fd-7c78-4b87-8728-eb515f3a851a","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"202cdea0-5891-48a6-8af1-80314a36eb55","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"3ac9664b-d4ce-4692-b5da-a7c10c93d1df","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"45654c9f-ca94-4a64-9fc8-fef9feaa45bc","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"4d1121ce-37e1-41aa-8c97-5768d1b3077f","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"528b17de-5785-42b8-b0cd-09167a33eaec","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"605514e0-9f84-4772-b4a1-815d66c32347","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"6e296b97-1391-45dc-8a34-7a009c036278","difficultyBand":"medium","calculatorUse":null,"totalMarks":8,"questionType":"LA","isCalculation":true},{"id":"70d48f46-9ed5-4abe-b540-298c6de7fef7","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"92b886f1-cacb-4682-a2ed-65fb97648785","difficultyBand":"medium","calculatorUse":null,"totalMarks":7,"questionType":"LA","isCalculation":true},{"id":"938b67ea-193f-465e-9d47-ae615215b1d3","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"e2928408-1275-4379-963a-b7641eeca420","difficultyBand":"hard","calculatorUse":null,"totalMarks":9,"questionType":"LA","isCalculation":true},{"id":"efaf632d-4adb-4ff8-ab51-b58991bc6672","difficultyBand":"hard","calculatorUse":null,"totalMarks":10,"questionType":"LA","isCalculation":true},{"id":"fb2f22a9-10c0-4724-b900-f806b261d2dc","difficultyBand":"medium","calculatorUse":null,"totalMarks":6,"questionType":"LA","isCalculation":true},{"id":"fd81699b-526f-4862-a6b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11.10 Solving Differential Equations

Question 8

11.10 Solving Differential Equations Questions

11.10 Solving Differential Equations

Question 8

11.10 Solving Differential Equations Questions