### Related Formula textEquation of tangent at (x, y): Y - y = fracdydx(X - x)$\text{Equation of tangent at } (x, y): Y - y = \frac{dy}{dx}(X - x)$ textArea of Right Triangle: frac12 times textbase times textheight$\text{Area of Right Triangle: } \frac{1}{2} \times \text{base} \times \text{height}$ ### Core Logic The line Y - y = y'(X - x)$Y - y = y'(X - x)$ intersects the axes to form a triangle. Let's find the intercepts: X-intercept (set Y=0$Y=0$): -y = y'(X - x) Rightarrow X = x - fracyy'$-y = y'(X - x) \Rightarrow X = x - \frac{y}{y'}$ Y-intercept (set X=0$X=0$): Y - y = y'(-x) Rightarrow Y = y - xy'$Y - y = y'(-x) \Rightarrow Y = y - xy'$ The area of the triangle bounded by the coordinate axes is: textArea = frac12 left( x - fracyy' right) (y - xy') = frac-y^22y' + 1$\text{Area} = \frac{1}{2} \left( x - \frac{y}{y'} \right) (y - xy') = \frac{-y^2}{2y'} + 1$

### Step 1: Setting up the Differential Equation Expand the Area equation: frac12 left[ xy - x^2y' - fracy^2y' + xy right] = frac-y^22y' + 1$\frac{1}{2} \left[ xy - x^2y' - \frac{y^2}{y'} + xy \right] = \frac{-y^2}{2y'} + 1$ Multiply by 2: 2xy - x^2y' - fracy^2y' = -fracy^2y' + 2$2xy - x^2y' - \frac{y^2}{y'} = -\frac{y^2}{y'} + 2$ Cancel -fracy^2y'$-\frac{y^2}{y'}$ from both sides: 2xy - x^2y' = 2$2xy - x^2y' = 2$ ### Step 2: Solving the Linear Differential Equation Rearrange into standard LDE form: x^2 fracdydx - 2xy = -2$x^2 \frac{dy}{dx} - 2xy = -2$ fracdydx - frac2xy = -frac2x^2$\frac{dy}{dx} - \frac{2}{x}y = -\frac{2}{x^2}$ Find the Integrating Factor (IF): textI.F. = e^int -frac2x dx = e^-2ln x = x^-2 = frac1x^2$\text{I.F.} = e^{\int -\frac{2}{x} dx} = e^{-2\ln x} = x^{-2} = \frac{1}{x^2}$ Multiply the LDE by the IF: y cdot frac1x^2 = int left( -frac2x^2 right) frac1x^2 dx + C$y \cdot \frac{1}{x^2} = \int \left( -\frac{2}{x^2} \right) \frac{1}{x^2} dx + C$ y x^-2 = -2 int x^-4 dx + C$y x^{-2} = -2 \int x^{-4} dx + C$ y x^-2 = -2 left( fracx^-3-3 right) + C$y x^{-2} = -2 \left( \frac{x^{-3}}{-3} \right) + C$ fracyx^2 = frac23x^3 + C$\frac{y}{x^2} = \frac{2}{3x^3} + C$ y = frac23x + Cx^2$y = \frac{2}{3x} + Cx^2$ ### Step 3: Applying Initial Conditions Given Y(1) = 1$Y(1) = 1$ (i.e., when x=1, y=1$x=1, y=1$): 1 = frac23(1) + C(1)^2 Rightarrow 1 = frac23 + C Rightarrow C = frac13$1 = \frac{2}{3(1)} + C(1)^2 \Rightarrow 1 = \frac{2}{3} + C \Rightarrow C = \frac{1}{3}$ Thus, the curve is: y = frac23x + fracx^23$y = \frac{2}{3x} + \frac{x^2}{3}$ We need to find 12Y(2)$12Y(2)$: Y(2) = frac23(2) + frac2^23 = frac13 + frac43 = frac53$Y(2) = \frac{2}{3(2)} + \frac{2^2}{3} = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$ 12Y(2) = 12 left( frac53 right) = 4 times 5 = 20$12Y(2) = 12 \left( \frac{5}{3} \right) = 4 \times 5 = 20$ ### 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

### Related Formula d(u cdot v) = u \, dv + v \, du$d(u \cdot v) = u \, dv + v \, du$ ### Core Logic Given the differential equation: sec x \, dy + \2(1 - x) tan x + (2x - x^2)\ \, dx = 0$\sec x \, dy + \{2(1 - x) \tan x + (2x - x^2)\} \, dx = 0$ Divide entirely by sec x$\sec x$ (which is 1/cos x$1/\cos x$): dy + \2(1 - x) sin x + (2x - x^2) cos x\ \, dx = 0$dy + \{2(1 - x) \sin x + (2x - x^2) \cos x\} \, dx = 0$ Rearranging it as an exact differential form: fracdydx = 2(x - 1) sin x + (x^2 - 2x) cos x$\frac{dy}{dx} = 2(x - 1) \sin x + (x^2 - 2x) \cos x$ ### Step 1: Integration using By-Parts Now integrate both sides: y(x) = int 2(x - 1) sin x \, dx + int (x^2 - 2x) cos x \, dx$y(x) = \int 2(x - 1) \sin x \, dx + \int (x^2 - 2x) \cos x \, dx$ Apply integration by parts on the second integral, where u = x^2 - 2x$u = x^2 - 2x$ and dv = cos x \, dx$dv = \cos x \, dx$: int (x^2 - 2x) cos x \, dx = (x^2 - 2x)(sin x) - int (2x - 2) sin x \, dx$\int (x^2 - 2x) \cos x \, dx = (x^2 - 2x)(\sin x) - \int (2x - 2) \sin x \, dx$ Substituting this back into y(x)$y(x)$: y(x) = int 2(x - 1) sin x \, dx + left[ (x^2 - 2x) sin x - int 2(x - 1) sin x \, dx right] + lambda$y(x) = \int 2(x - 1) \sin x \, dx + \left[ (x^2 - 2x) \sin x - \int 2(x - 1) \sin x \, dx \right] + \lambda$ The integrals cancel out perfectly: y(x) = (x^2 - 2x) sin x + lambda$y(x) = (x^2 - 2x) \sin x + \lambda$ ### Step 2: Finding Constant and Final Value Use the initial condition y(0) = 2$y(0) = 2$: y(0) = 0 + lambda Rightarrow lambda = 2$y(0) = 0 + \lambda \Rightarrow \lambda = 2$ Thus, the solution is: y(x) = (x^2 - 2x) sin x + 2$y(x) = (x^2 - 2x) \sin x + 2$ To find y(2)$y(2)$: y(2) = (2^2 - 2(2)) sin 2 + 2 = (4 - 4) sin 2 + 2 = 2$y(2) = (2^2 - 2(2)) \sin 2 + 2 = (4 - 4) \sin 2 + 2 = 2$ ### Pattern Recognition When an integrand is a mix of polynomial and trigonometric terms, grouping them to form an exact differential or observing that integration by parts on one term perfectly cancels the other term is a classic structural trick. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Integrals

### Related Formula For a linear differential equation fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$: IF = e^int P(x) dx$IF = e^{\int P(x) dx}$ Solution: y cdot IF = int Q(x) cdot IF \, dx + C$y \cdot IF = \int Q(x) \cdot IF \, dx + C$ ### Core Logic Rearrange the given differential equation into standard linear form: (1 - x^2) fracdydx = xy + (x^3 + 2)sqrt3(1 - x^2)$(1 - x^2) \frac{dy}{dx} = xy + (x^3 + 2)\sqrt{3(1 - x^2)}$ fracdydx - fracx1 - x^2 y = frac(x^3 + 2)sqrt3(1 - x^2)1 - x^2$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{(x^3 + 2)\sqrt{3(1 - x^2)}}{1 - x^2}$ fracdydx - fracx1 - x^2 y = fracsqrt3(x^3 + 2)sqrt1 - x^2$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}}$ Identify P(x) = -fracx1 - x^2$P(x) = -\frac{x}{1 - x^2}$. ### Step 1: Finding Integrating Factor (IF) IF = e^int -fracx1 - x^2 dx$IF = e^{\int -\frac{x}{1 - x^2} dx}$ Let 1 - x^2 = t Rightarrow -2x dx = dt Rightarrow -x dx = fracdt2$1 - x^2 = t \Rightarrow -2x dx = dt \Rightarrow -x dx = \frac{dt}{2}$. IF = e^frac12 int frac1t dt = e^frac12 ln t = e^ln sqrtt = sqrt1 - x^2$IF = e^{\frac{1}{2} \int \frac{1}{t} dt} = e^{\frac{1}{2} \ln t} = e^{\ln \sqrt{t}} = \sqrt{1 - x^2}$ ### Step 2: General Solution The solution is given by: y sqrt1 - x^2 = int left( fracsqrt3(x^3 + 2)sqrt1 - x^2 right) sqrt1 - x^2 \, dx + C$y \sqrt{1 - x^2} = \int \left( \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}} \right) \sqrt{1 - x^2} \, dx + C$ y sqrt1 - x^2 = sqrt3 int (x^3 + 2) \, dx + C$y \sqrt{1 - x^2} = \sqrt{3} \int (x^3 + 2) \, dx + C$ y sqrt1 - x^2 = sqrt3 left( fracx^44 + 2x right) + C$y \sqrt{1 - x^2} = \sqrt{3} \left( \frac{x^4}{4} + 2x \right) + C$ ### Step 3: Finding Constant C Using y(0) = 0$y(0) = 0$: 0 cdot 1 = sqrt3(0 + 0) + C Rightarrow C = 0$0 \cdot 1 = \sqrt{3}(0 + 0) + C \Rightarrow C = 0$ So, y(x) = fracsqrt3sqrt1 - x^2 left( fracx^44 + 2x right)$y(x) = \frac{\sqrt{3}}{\sqrt{1 - x^2}} \left( \frac{x^4}{4} + 2x \right)$. ### Step 4: Evaluating required point Substitute x = 1/2$x = 1/2$: yleft(frac12right) = fracsqrt3sqrt1 - 1/4 left( frac1/164 + 2left(frac12right) right)$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{\sqrt{1 - 1/4}} \left( \frac{1/16}{4} + 2\left(\frac{1}{2}\right) \right)$ Wait, frac(1/2)^44 = frac1/164 = frac164$\frac{(1/2)^4}{4} = \frac{1/16}{4} = \frac{1}{64}$. Let's recheck the expression: yleft(frac12right) = fracsqrt3sqrt3/4 left( frac164 + 1 right)$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{\sqrt{3/4}} \left( \frac{1}{64} + 1 \right)$ = fracsqrt3sqrt3/2 left( frac6564 right) = 2 times frac6564 = frac6532$= \frac{\sqrt{3}}{\sqrt{3}/2} \left( \frac{65}{64} \right) = 2 \times \frac{65}{64} = \frac{65}{32}$ Here, m = 65$m = 65$ and n = 32$n = 32$. They are co-prime. Thus, m + n = 65 + 32 = 97$m + n = 65 + 32 = 97$. ### Pattern Recognition Whenever roots matching the integration denominator appear on the RHS of a linear differential setup, it is a high-confidence signal that the integrating factor cleanly annihilates the fractional root component during the solution stage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Related Formula int fracdTT-T_s = -K int dt implies ln|T-T_s| = -Kt + C$\int \frac{dT}{T-T_s} = -K \int dt \implies \ln|T-T_s| = -Kt + C$ ### Core Logic Given fracdTdt = -K(T-80)$\frac{dT}{dt} = -K(T-80)$. Integrating from t=0$t=0$ to t$t$: int_160^T fracdTT-80 = -K int_0^t dt$\int_{160}^{T} \frac{dT}{T-80} = -K \int_0^t dt$ [ln|T-80|]_160^T = -Kt$[\ln|T-80|]_{160}^T = -Kt$ lnleft(fracT-8080right) = -Kt implies T(t) = 80 + 80e^-Kt$\ln\left(\frac{T-80}{80}\right) = -Kt \implies T(t) = 80 + 80e^{-Kt}$ Given T(15) = 120$T(15) = 120$: 120 = 80 + 80e^-15K implies 40 = 80e^-15K implies e^-15K = frac12$120 = 80 + 80e^{-15K} \implies 40 = 80e^{-15K} \implies e^{-15K} = \frac{1}{2}$ To find T(45)$T(45)$: T(45) = 80 + 80e^-45K = 80 + 80(e^-15K)^3$T(45) = 80 + 80e^{-45K} = 80 + 80(e^{-15K})^3$ = 80 + 80left(frac12right)^3$= 80 + 80\left(\frac{1}{2}\right)^3$ = 80 + 80left(frac18right) = 90$= 80 + 80\left(\frac{1}{8}\right) = 90$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations