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JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)

Let Y = Y(X) be a curve lying in the first quadrant such that the area enclosed by the line Y - y = Y'(x)(X - x) and the co-ordinate axes, where (x, y) is any point on the curve, is always frac-y^22Y'(x) + 1 , Y'(x) neq 0 . If Y(1) = 1 , then 12Y(2) equals

Numerical Answer Type:
Enter a numerical value Answer: 20 to 20 +4 marks

Solution & Explanation

### Related Formula textEquation of tangent at (x, y): Y - y = fracdydx(X - x) textArea of Right Triangle: frac12 times textbase times textheight ### Core Logic The line Y - y = y'(X - x) intersects the axes to form a triangle. Let's find the intercepts: X-intercept (set Y=0): -y = y'(X - x) Rightarrow X = x - fracyy' Y-intercept (set X=0): Y - y = y'(-x) Rightarrow Y = y - xy' The area of the triangle bounded by the coordinate axes is: textArea = frac12 left( x - fracyy' right) (y - xy') = frac-y^22y' + 1
Linear Differential Equation diagram for Q21 - JEE Main 2024 Evening
Linear Differential Equation diagram for Q21 - JEE Main 2024 Evening
### Step 1: Setting up the Differential Equation Expand the Area equation: frac12 left[ xy - x^2y' - fracy^2y' + xy right] = frac-y^22y' + 1 Multiply by 2: 2xy - x^2y' - fracy^2y' = -fracy^2y' + 2 Cancel -fracy^2y' from both sides: 2xy - x^2y' = 2 ### Step 2: Solving the Linear Differential Equation Rearrange into standard LDE form: x^2 fracdydx - 2xy = -2 fracdydx - frac2xy = -frac2x^2 Find the Integrating Factor (IF): textI.F. = e^int -frac2x dx = e^-2ln x = x^-2 = frac1x^2 Multiply the LDE by the IF: y cdot frac1x^2 = int left( -frac2x^2 right) frac1x^2 dx + C y x^-2 = -2 int x^-4 dx + C y x^-2 = -2 left( fracx^-3-3 right) + C fracyx^2 = frac23x^3 + C y = frac23x + Cx^2 ### Step 3: Applying Initial Conditions Given Y(1) = 1 (i.e., when x=1, y=1): 1 = frac23(1) + C(1)^2 Rightarrow 1 = frac23 + C Rightarrow C = frac13 Thus, the curve is: y = frac23x + fracx^23 We need to find 12Y(2): Y(2) = frac23(2) + frac2^23 = frac13 + frac43 = frac53 12Y(2) = 12 left( frac53 right) = 4 times 5 = 20 ### Pattern Recognition Geometrical word problems forming an Area differential equation almost always reduce to a first-order Linear Differential Equation (LDE) or an Exact differential form once you cleanly substitute the axis intercepts. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Application of Derivatives

Reference Study Guides

Differential Equations

More Differential Equations Previous-Year Questions — Page 2

Q67 jee_main_2025_08_april_evening Linear Differential Equations
Let mathrmf(x) = x - 1 and mathrmg(x) = e^x for mathbfxin mathbbR. If fracmathrmdymathrmdx = left(mathrme^-2sqrtmathrmx mathrmgBig(mathrmfbig(mathrmf(mathrmx)big)Big) - fracmathrmysqrtmathrmxright), mathrmy(0) = 0, then mathrmy(1) is:
  • A. frac1 - mathrme^2mathrme^4
  • B. frac2mathrme - 1mathrme^3
  • C. fracmathrme - 1mathrme^4
  • D. frac1 - mathrme^3mathrme^4

Solution

### Related Formula textLinear Form: fracdydx + P(x)y = Q(x) implies I.F. = e^int P(x) \, dx ### Core Logic Evaluate composite function layers to organize equation segments into a standard first-order linear differential form, then introduce proper scaling factors. ### Step 1: Simplify Composite Functional Core f(f(x)) = (x-1) - 1 = x - 2 implies g(f(f(x))) = e^x-2 ### Step 2: Restructure Equation and Compute Integrating Factor fracdydx + frac1sqrtxy = e^-2sqrtx cdot e^x-2 = e^x - 2sqrtx - 2 I.F. = e^int frac1sqrtx \, dx = e^2sqrtx ### Step 3: Integrate General Tracking Steps y times e^2sqrtx = int e^2sqrtx times e^x - 2sqrtx - 2 \, dx + c = int e^x-2 \, dx + c y times e^2sqrtx = e^x-2 + c Using boundary values x=0, y=0 implies 0 = e^-2 + c implies c = -e^-2. ### Step 4: Evaluate Value Bounds At Point Profile y times e^2sqrtx = e^x-2 - e^-2 At x = 1: y(1) times e^2 = e^-1 - e^-2 implies y(1) = frace^-1 - e^-2e^2 = frace-1e^4 ### Pattern Recognition Composite layouts often produce exponent segments designed to cancel tracking multiples within integrating factor components automatically. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 11 Mathematics: Relations and Functions
Q70 jee_main_2025_29_jan_evening Linear Differential Equations
If for the solution curve y = f(x) of the differential equation fracdydx + (tan x)y = frac2 + sec x(1 + 2sec x)^2, x in left(frac-pi2, fracpi2right), fleft(fracpi3right) = fracsqrt310, then fleft(fracpi4right) is equal to:
  • A. frac9sqrt3 + 310(4 + sqrt3)
  • B. fracsqrt3 + 110left(4 + sqrt3right)
  • C. frac5 - sqrt32sqrt2
  • D. frac4 - sqrt214

Solution

### Related Formula Integrating factor (I.F.) for a linear differential equation fracdydx + Py = Q: textI.F. = e^int P \, dx General solution: y cdot (textI.F.) = int Q cdot (textI.F.) \, dx ### Core Logic Given P = tan x, compute Integrating Factor: textI.F. = e^int tan x \, dx = e^ln(sec x) = sec x Set up integrated expression solution layout: y cdot sec x = int frac2 + sec x(1 + 2sec x)^2 cdot sec x \, dx = int frac2cos x + 1(cos x + 2)^2 cdot dx ### Step 1: Evaluate Integration with Half-Angle Substitutions Using tangent half-angle substitution t = tanfracx2 transformations simplifies the integral loop structure down to: y cdot sec x = frac2t + frac3t + C Plugging entry condition parameters fleft(fracpi3 ight) = fracsqrt310 tracking t = frac1sqrt3 explicitly isolates boundary condition constant C: C = 0 ### Step 2: Calculate Target Point Value At target query point x = fracpi4, half-angle parameters scale to t = sqrt2 - 1: y cdot sqrt2 = frac2sqrt2 - 1 + frac3sqrt2 - 1 = frac2(sqrt2 - 1)6 - 2sqrt2 y = frac4 - sqrt214 ### Pattern Recognition When integrating complex rational expressions involving trigonometric values, half-angle substitution methods (t = tanfracx2) are standard for reducing polynomial degrees. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations
Q66 jee_main_2025_28_jan_morning Leibniz Rule and Linear Differential Equations
Let for some function y = f(x), int_0^x t f(t) dt = x^2 f(x), x > 0 text and f(2) = 3. Then f(6) is equal to: (1) 1 (2) 2 (3) 6 (4) 3
  • A. 1
  • B. 2
  • C. 6
  • D. 3

Solution

### Related Formula Leibniz Rule for differentiating under the integral sign: fracddx left[ int_psi(x)^phi(x) f(t) dt right] = f(phi(x))phi^prime(x) - f(psi(x))psi^prime(x) ### Core Logic Differentiate both sides of the integral equation with respect to x: xf(x) = x^2 f^prime(x) + 2xf(x) implies -xf(x) = x^2 f^prime(x) ### Step 1: Solving the Separable Differential Equation Separating variables: int fracf^prime(x)f(x) dx = int -frac1x dx implies ln |f(x)| = -ln x + ln c implies f(x) = fraccx ### Step 2: Applying Boundary Constraints Given f(2) = 3: 3 = fracc2 implies c = 6 implies f(x) = frac6x Evaluating for x = 6: f(6) = frac66 = 1 ### Pattern Recognition Differentiating integral statements instantly converts complex integral equations into clean, separable differential equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations
Q62 jee_main_2025_03_april_morning Linear Differential Equations
Let g be a differentiable function such that int_0^xg(t)dt=x-int_0^xtg(t)dt [cite: 568], xge0 [cite: 569] and let y=y(x) satisfy the differential equation fracdydx - ytan x = 2(x+1)sec x \, g(x) [cite: 571, 575, 578, 581], xin[0,fracpi2)[cite: 581]. If y(0)=0 [cite: 579] then yleft(fracpi3right) is equal to[cite: 582]:
  • A. frac2pi3sqrt3
  • B. frac4pi3
  • C. frac2pi3
  • D. frac4pi3sqrt3

Solution

### Related Formula Leibniz Integral Rule for differentiation: fracmathrmdmathrmdxleft(int_0^x f(t)mathrmdtright) = f(x) ### Core Logic Differentiate the given integral relation using Leibniz rule [cite: 1344]: fracmathrmdmathrmdxleft[int_0^xg(t)dtright] = fracmathrmdmathrmdxleft[x-int_0^xtg(t)dtright] [cite: 1344] g(x) = 1 - xg(x) implies g(x)(1+x) = 1 implies g(x) = frac11+x [cite: 1345] Substitute g(x) into target differential equation configuration [cite: 1346]: fracmathrmdymathrmdx - ytan x = 2(x+1)sec x cdot left(frac11+xright) = 2sec x [cite: 1346] ### Step 1: Finding the Integrating Factor This matches a linear form fracmathrmdymathrmdx + P(x)y = Q(x) where P(x) = -tan x. textI.F. = e^int -tan x \, mathrmdx = e^ln|cos x| = cos x [cite: 1346] Write general functional solution template [cite: 1348]: y cdot cos x = int (2sec x cdot cos x) \, mathrmdx = int 2 \, mathrmdx = 2x + C [cite: 1348] Given boundary condition y(0) = 0 implies 0 = 0 + C implies C = 0 [cite: 1349]. y(x) = frac2xcos x = 2xsec x [cite: 1350] ### Step 2: Numeric substitution Substitute variable parameter values x = fracpi3 [cite: 1352]: yleft(fracpi3right) = 2left(fracpi3right)secleft(fracpi3right) = frac2pi3 cdot 2 = frac4pi3 [cite: 1351] ### Pattern Recognition Integral functional definitions are codes for simpler underlying derivatives. Applying Leibniz rule immediately extracts the true variable functions. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations
Q66 jee_main_2025_04_april_evening Linear Differential Equations
If a curve y = y(x) passes through the point left(1, fracpi2right) and satisfies the differential equation (7x^4 cot y - e^x mathrmcosec\,y) fracdxdy = x^5, x geq 1, then at x = 2, the value of cosine is:
  • A. frac2mathrme^2 - mathrme64
  • B. frac2mathrme^2 + mathrme64
  • C. frac2mathrme^2 - mathrme128
  • D. frac2mathrme^2 + mathrme128

Solution

### Core Logic Let's rearrange the given differential equation by expressing fracdydx: x^5 fracdydx = 7x^4 cot y - e^x csc y Dividing both sides by x^5: fracdydx = frac7x cot y - frace^xx^5 csc y Multiply the entire equation by sin y to clear denominators: sin y fracdydx - frac7x cos y = -frace^xx^5 This can be transformed into a linear form by substituting t = -cos y. Then fracdtdx = sin y fracdydx. ### Step 1: Solving the Linear ODE Substituting t leads to: fracdtdx + frac7x t = -frace^xx^5 This is a standard linear first-order ODE with P(x) = frac7x. The Integrating Factor (I.F.) is: textI.F. = e^int frac7x dx = e^7 ln x = x^7 The general solution is: t cdot x^7 = int left(-frace^xx^5right) cdot x^7 dx = -int x^2 e^x dx ### Step 2: Evaluating the Integration and Constant Using integration by parts for int x^2 e^x dx: int x^2 e^x dx = x^2 e^x - 2xe^x + 2e^x Substituting this back: -cos y cdot x^7 = -e^x(x^2 - 2x + 2) + C cos y cdot x^7 = e^x(x^2 - 2x + 2) - C Since the curve passes through left(1, fracpi2right): cosleft(fracpi2right) cdot (1)^7 = e^1(1^2 - 2(1) + 2) - C implies 0 = e(1) - C implies C = e ### Step 3: Calculating cos y at x = 2 Now substitute x = 2 and C = e into our equation block: cos y cdot (2^7) = e^2(2^2 - 2(2) + 2) - e cos y cdot 128 = e^2(4 - 4 + 2) - e = 2e^2 - e cos y = frac2e^2 - e128 ### Pattern Recognition When trigonometric terms are mixed inside an ODE containing derivative blocks like fracdxdy or fracdydx, check if clearing denominators using sin y or cos y reveals a standard substitution path for a Linear ODE. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

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