### Related Formula textLinear Form: fracdydx + P(x)y = Q(x) implies I.F. = e^int P(x) \, dx$\text{Linear Form: } \frac{dy}{dx} + P(x)y = Q(x) \implies I.F. = e^{\int P(x) \, dx}$ ### 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$f(f(x)) = (x-1) - 1 = x - 2 \implies g(f(f(x))) = e^{x-2}$ ### Step 2: Restructure Equation and Compute Integrating Factor fracdydx + frac1sqrtxy = e^-2sqrtx cdot e^x-2 = e^x - 2sqrtx - 2$\frac{dy}{dx} + \frac{1}{\sqrt{x}}y = e^{-2\sqrt{x}} \cdot e^{x-2} = e^{x - 2\sqrt{x} - 2}$ I.F. = e^int frac1sqrtx \, dx = e^2sqrtx$I.F. = e^{\int \frac{1}{\sqrt{x}} \, dx} = e^{2\sqrt{x}}$ ### 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$y \times e^{2\sqrt{x}} = \int e^{2\sqrt{x}} \times e^{x - 2\sqrt{x} - 2} \, dx + c = \int e^{x-2} \, dx + c$ y times e^2sqrtx = e^x-2 + c$y \times e^{2\sqrt{x}} = e^{x-2} + c$ Using boundary values x=0, y=0 implies 0 = e^-2 + c implies c = -e^-2$x=0, y=0 \implies 0 = e^{-2} + c \implies c = -e^{-2}$. ### Step 4: Evaluate Value Bounds At Point Profile y times e^2sqrtx = e^x-2 - e^-2$y \times e^{2\sqrt{x}} = e^{x-2} - e^{-2}$ At x = 1$x = 1$: y(1) times e^2 = e^-1 - e^-2 implies y(1) = frace^-1 - e^-2e^2 = frace-1e^4$y(1) \times e^{2} = e^{-1} - e^{-2} \implies y(1) = \frac{e^{-1} - e^{-2}}{e^2} = \frac{e-1}{e^4}$ ### 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

### Related Formula Integrating factor (I.F.) for a linear differential equation fracdydx + Py = Q$\frac{dy}{dx} + Py = Q$: textI.F. = e^int P \, dx$\text{I.F.} = e^{\int P \, dx}$ General solution: y cdot (textI.F.) = int Q cdot (textI.F.) \, dx$y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) \, dx$ ### Core Logic Given P = tan x$P = \tan x$, compute Integrating Factor: textI.F. = e^int tan x \, dx = e^ln(sec x) = sec x$\text{I.F.} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x$ 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$y \cdot \sec x = \int \frac{2 + \sec x}{(1 + 2\sec x)^2} \cdot \sec x \, dx = \int \frac{2\cos x + 1}{(\cos x + 2)^2} \cdot dx$ ### Step 1: Evaluate Integration with Half-Angle Substitutions Using tangent half-angle substitution t = tanfracx2$t = \tan\frac{x}{2}$ transformations simplifies the integral loop structure down to: y cdot sec x = frac2t + frac3t + C$y \cdot \sec x = \frac{2}{t + \frac{3}{t}} + C$ Plugging entry condition parameters fleft(fracpi3 ight) = fracsqrt310$f\left(\frac{\pi}{3} ight) = \frac{\sqrt{3}}{10}$ tracking t = frac1sqrt3$t = \frac{1}{\sqrt{3}}$ explicitly isolates boundary condition constant C$C$: C = 0$C = 0$ ### Step 2: Calculate Target Point Value At target query point x = fracpi4$x = \frac{\pi}{4}$, half-angle parameters scale to t = sqrt2 - 1$t = \sqrt{2} - 1$: y cdot sqrt2 = frac2sqrt2 - 1 + frac3sqrt2 - 1 = frac2(sqrt2 - 1)6 - 2sqrt2$y \cdot \sqrt{2} = \frac{2}{\sqrt{2} - 1 + \frac{3}{\sqrt{2} - 1}} = \frac{2(\sqrt{2} - 1)}{6 - 2\sqrt{2}}$ y = frac4 - sqrt214$y = \frac{4 - \sqrt{2}}{14}$ ### Pattern Recognition When integrating complex rational expressions involving trigonometric values, half-angle substitution methods (t = tanfracx2$t = \tan\frac{x}{2}$) are standard for reducing polynomial degrees. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### 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)$\frac{d}{dx} \left[ \int_{\psi(x)}^{\phi(x)} f(t) dt \right] = f(\phi(x))\phi^\prime(x) - f(\psi(x))\psi^\prime(x)$ ### Core Logic Differentiate both sides of the integral equation with respect to x$x$: xf(x) = x^2 f^prime(x) + 2xf(x) implies -xf(x) = x^2 f^prime(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$\int \frac{f^\prime(x)}{f(x)} dx = \int -\frac{1}{x} dx \implies \ln |f(x)| = -\ln x + \ln c \implies f(x) = \frac{c}{x}$ ### Step 2: Applying Boundary Constraints Given f(2) = 3$f(2) = 3$: 3 = fracc2 implies c = 6 implies f(x) = frac6x$3 = \frac{c}{2} \implies c = 6 \implies f(x) = \frac{6}{x}$ Evaluating for x = 6$x = 6$: f(6) = frac66 = 1$f(6) = \frac{6}{6} = 1$ ### 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

### Related Formula Leibniz Integral Rule for differentiation: fracmathrmdmathrmdxleft(int_0^x f(t)mathrmdtright) = f(x)$\frac{\mathrm{d}}{\mathrm{d}x}\left(\int_0^x f(t)\mathrm{d}t\right) = f(x)$ ### 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]$\frac{\mathrm{d}}{\mathrm{d}x}\left[\int_{0}^{x}g(t)dt\right] = \frac{\mathrm{d}}{\mathrm{d}x}\left[x-\int_{0}^{x}tg(t)dt\right]$ [cite: 1344] g(x) = 1 - xg(x) implies g(x)(1+x) = 1 implies g(x) = frac11+x$g(x) = 1 - xg(x) \implies g(x)(1+x) = 1 \implies g(x) = \frac{1}{1+x}$ [cite: 1345] Substitute g(x)$g(x)$ into target differential equation configuration [cite: 1346]: fracmathrmdymathrmdx - ytan x = 2(x+1)sec x cdot left(frac11+xright) = 2sec x$\frac{\mathrm{d}y}{\mathrm{d}x} - y\tan x = 2(x+1)\sec x \cdot \left(\frac{1}{1+x}\right) = 2\sec x$ [cite: 1346] ### Step 1: Finding the Integrating Factor This matches a linear form fracmathrmdymathrmdx + P(x)y = Q(x)$\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)$ where P(x) = -tan x$P(x) = -\tan x$. textI.F. = e^int -tan x \, mathrmdx = e^ln|cos x| = cos x$\text{I.F.} = e^{\int -\tan x \, \mathrm{d}x} = e^{\ln|\cos x|} = \cos x$ [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$y \cdot \cos x = \int (2\sec x \cdot \cos x) \, \mathrm{d}x = \int 2 \, \mathrm{d}x = 2x + C$ [cite: 1348] Given boundary condition y(0) = 0 implies 0 = 0 + C implies C = 0$y(0) = 0 \implies 0 = 0 + C \implies C = 0$ [cite: 1349]. y(x) = frac2xcos x = 2xsec x$y(x) = \frac{2x}{\cos x} = 2x\sec x$ [cite: 1350] ### Step 2: Numeric substitution Substitute variable parameter values x = fracpi3$x = \frac{\pi}{3}$ [cite: 1352]: yleft(fracpi3right) = 2left(fracpi3right)secleft(fracpi3right) = frac2pi3 cdot 2 = frac4pi3$y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} \cdot 2 = \frac{4\pi}{3}$ [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

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