Let y = y(x)$y = y(x)$ be the solution of the differential equation fracdydx + 3(tan^2 x)y + 3y = sec^2 x, y(0) = frac13 + e^3$\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x, y(0) = \frac{1}{3} + e^3$. Then yleft(fracpi4 ight)$y\left(\frac{pi}{4} ight)$ is equal to

### Related Formula textStandard Form of LDE: fracdydx + P(x)y = Q(x)$$\text{Standard Form of LDE: } \frac{dy}{dx} + P(x)y = Q(x)$$ I.F. = e^int P(x)dx$$I.F. = e^{\int P(x)dx}$$ textSolution: y cdot (I.F.) = int Q(x) cdot (I.F.) dx + C$$\text{Solution: } y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$ ### Core Logic Given the differential equation: ysin 2x + sin x - (1+cos^2 x)fracdydx = 0$$y\sin 2x + \sin x - (1+\cos^2 x)\frac{dy}{dx} = 0$$ Rearrange it into the standard linear differential equation format: (1+cos^2 x)fracdydx - ysin 2x = sin x$$(1+\cos^2 x)\frac{dy}{dx} - y\sin 2x = \sin x$$ fracdydx - left(fracsin 2x1+cos^2 xright)y = fracsin x1+cos^2 x$$\frac{dy}{dx} - \left(\frac{\sin 2x}{1+\cos^2 x}\right)y = \frac{\sin x}{1+\cos^2 x}$$ ### Step 1: Find the Integrating Factor (I.F.) Here, P(x) = -fracsin 2x1+cos^2 x$P(x) = -\frac{\sin 2x}{1+\cos^2 x}$. I.F. = e^int -fracsin 2x1+cos^2 x dx$$I.F. = e^{\int -\frac{\sin 2x}{1+\cos^2 x} dx}$$ Let 1+cos^2 x = t$1+\cos^2 x = t$, then dt = -2cos xsin x \, dx = -sin 2x \, dx$dt = -2\cos x\sin x \, dx = -\sin 2x \, dx$. I.F. = e^int frac1t dt = e^ln t = t$$I.F. = e^{\int \frac{1}{t} dt} = e^{\ln t} = t$$ I.F. = 1+cos^2 x$$I.F. = 1+\cos^2 x$$ ### Step 2: Solve the Differential Equation Multiply the LDE by the integrating factor: y cdot (1+cos^2 x) = int left(fracsin x1+cos^2 xright) cdot (1+cos^2 x) dx$$y \cdot (1+\cos^2 x) = \int \left(\frac{\sin x}{1+\cos^2 x}\right) \cdot (1+\cos^2 x) dx$$ y(1+cos^2 x) = int sin x dx$$y(1+\cos^2 x) = \int \sin x dx$$ y(1+cos^2 x) = -cos x + C$$y(1+\cos^2 x) = -\cos x + C$$ Apply the initial condition f(0) = 0$f(0) = 0$ (which means y=0$y=0$ when x=0$x=0$): 0 cdot (1+1) = -1 + C Rightarrow C = 1$$0 \cdot (1+1) = -1 + C \Rightarrow C = 1$$ The particular solution is: y(1+cos^2 x) = 1 - cos x$$y(1+\cos^2 x) = 1 - \cos x$$ ### Step 3: Evaluate at x = pi/2 To find f(fracpi2)$f(\frac{\pi}{2})$, substitute x = fracpi2$x = \frac{\pi}{2}$: yleft(1+cos^2left(fracpi2right)right) = 1 - cosleft(fracpi2right)$$y\left(1+\cos^2\left(\frac{\pi}{2}\right)\right) = 1 - \cos\left(\frac{\pi}{2}\right)$$ y(1+0) = 1 - 0$y(1+0) = 1 - 0$ y = 1$y = 1$ ### Pattern Recognition Whenever y$y$ and y'$y'$ appear linearly separated by complex trigonometric polynomials, format aggressively into standard LDE form. The derivative of 1+cos^2x$1+\cos^2x$ is exactly -sin 2x$-\sin 2x$, making the logarithmic integration seamless. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integral Calculus

### Related Formula int frac11+y^2 dy = tan^-1y + C$$\int \frac{1}{1+y^2} dy = \tan^{-1}y + C$$ int f(x) f'(x) dx = frac[f(x)]^22 + C$$\int f(x) f'(x) dx = \frac{[f(x)]^2}{2} + C$$ tan(A - B) = fractan A - tan B1 + tan A tan B$$\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$ ### Core Logic Rearrange the given differential equation to separate variables x$x$ and y$y$: (1+y^2)(1+log_e x)dx + x dy = 0$$(1+y^2)(1+\log_e x)dx + x dy = 0$$ frac1+log_e xx dx + frac11+y^2 dy = 0$$\frac{1+\log_e x}{x} dx + \frac{1}{1+y^2} dy = 0$$ Integrate both sides: int frac1+log_e xx dx + int frac11+y^2 dy = C$$\int \frac{1+\log_e x}{x} dx + \int \frac{1}{1+y^2} dy = C$$ ### Step 1: Execute Integration For the first integral, let u = 1+log_e x$u = 1+\log_e x$, so du = frac1x dx$du = \frac{1}{x} dx$. int u \, du = fracu^22 = frac(1+log_e x)^22$$\int u \, du = \frac{u^2}{2} = \frac{(1+\log_e x)^2}{2}$$ For the second integral: int frac11+y^2 dy = tan^-1y$$\int \frac{1}{1+y^2} dy = \tan^{-1}y$$ Putting them together: frac(1+log_e x)^22 + tan^-1y = C$$\frac{(1+\log_e x)^2}{2} + \tan^{-1}y = C$$ ### Step 2: Apply Boundary Conditions The curve passes through (1, 1)$(1, 1)$. Substitute x=1, y=1$x=1, y=1$: frac(1+log_e 1)^22 + tan^-1(1) = C$$\frac{(1+\log_e 1)^2}{2} + \tan^{-1}(1) = C$$ Since log_e 1 = 0$\log_e 1 = 0$ and tan^-1(1) = fracpi4$\tan^{-1}(1) = \frac{\pi}{4}$: frac1^22 + fracpi4 = C Rightarrow C = frac12 + fracpi4$$\frac{1^2}{2} + \frac{\pi}{4} = C \Rightarrow C = \frac{1}{2} + \frac{\pi}{4}$$ Wait, the solution rearranges differently: int (frac1x + fracln xx) dx = ln x + frac(ln x)^22$\int (\frac{1}{x} + \frac{\ln x}{x}) dx = \ln x + \frac{(\ln x)^2}{2}$. Let's use this to stay fully synced with the PDF's algebraic step. ln x + frac(ln x)^22 + tan^-1 y = C$$\ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = C$$ At (1,1)$(1,1)$: ln(1) + frac(ln 1)^22 + tan^-1(1) = C Rightarrow 0 + 0 + fracpi4 = C Rightarrow C = fracpi4$$\ln(1) + \frac{(\ln 1)^2}{2} + \tan^{-1}(1) = C \Rightarrow 0 + 0 + \frac{\pi}{4} = C \Rightarrow C = \frac{\pi}{4}$$ The general equation simplifies nicely to: ln x + frac(ln x)^22 + tan^-1 y = fracpi4$$\ln x + \frac{(\ln x)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$$ ### Step 3: Evaluate at x = e Substitute x=e$x=e$ into the equation to find y(e)$y(e)$: ln e + frac(ln e)^22 + tan^-1 y = fracpi4$$\ln e + \frac{(\ln e)^2}{2} + \tan^{-1} y = \frac{\pi}{4}$$ 1 + frac1^22 + tan^-1 y = fracpi4$$1 + \frac{1^2}{2} + \tan^{-1} y = \frac{\pi}{4}$$ frac32 + tan^-1 y = fracpi4$$\frac{3}{2} + \tan^{-1} y = \frac{\pi}{4}$$ tan^-1 y = fracpi4 - frac32$$\tan^{-1} y = \frac{\pi}{4} - \frac{3}{2}$$ y = tanleft(fracpi4 - frac32right)$$y = \tan\left(\frac{\pi}{4} - \frac{3}{2}\right)$$ Using the compound angle formula tan(A-B)$\tan(A-B)$: y = fractan(pi/4) - tan(3/2)1 + tan(pi/4)tan(3/2) = frac1 - tan(3/2)1 + tan(3/2)$$y = \frac{\tan(\pi/4) - \tan(3/2)}{1 + \tan(\pi/4)\tan(3/2)} = \frac{1 - \tan(3/2)}{1 + \tan(3/2)}$$ Comparing this format to y(e) = fracalpha-tan(3/2)beta+tan(3/2)$y(e) = \frac{\alpha-\tan(3/2)}{\beta+\tan(3/2)}$ gives: alpha = 1, quad beta = 1$$\alpha = 1, \quad \beta = 1$$ Calculate alpha + 2beta$\alpha + 2\beta$: 1 + 2(1) = 3$1 + 2(1) = 3$ ### Pattern Recognition For expressions involving separated x$x$ and y$y$ logarithmic terms, always split int frac1+ln xxdx$\int \frac{1+\ln x}{x}dx$ into int frac1x + int fracln xx$\int \frac{1}{x} + \int \frac{\ln x}{x}$ to avoid carrying a non-zero shift frac12$\frac{1}{2}$ into the constant C$C$. This allows direct identity matching for tangent sums. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 11 Mathematics: Trigonometric Functions

### Related Formula textChain rule for partial differentiation: fracpartialpartial x fleft(fracxyright) = f'left(fracxyright) cdot frac1y$$\text{Chain rule for partial differentiation: } \frac{\partial}{\partial x} f\left(\frac{x}{y}\right) = f'\left(\frac{x}{y}\right) \cdot \frac{1}{y}$$ ### Core Logic Given functional equation: fleft(fracxyright) = fracf(x)f(y)$f\left(\frac{x}{y}\right) = \frac{f(x)}{f(y)}$ First, put x=1, y=1$x=1, y=1$: f(1) = fracf(1)f(1) Rightarrow f(1) = 1$f(1) = \frac{f(1)}{f(1)} \Rightarrow f(1) = 1$ (since f(y) neq 0$f(y) \neq 0$ implies f(1) neq 0$f(1) \neq 0$). ### Step 1: Partial Differentiation Partially differentiate the given equation with respect to x$x$, keeping y$y$ constant: f'left(fracxyright) cdot frac1y = frac1f(y) cdot f'(x)$$f'\left(\frac{x}{y}\right) \cdot \frac{1}{y} = \frac{1}{f(y)} \cdot f'(x)$$ Let x to y$x \to y$ so that fracxy to 1$\frac{x}{y} \to 1$. Wait, let's substitute y = x$y = x$ directly: f'(1) cdot frac1x = frac1f(x) cdot f'(x)$$f'(1) \cdot \frac{1}{x} = \frac{1}{f(x)} \cdot f'(x)$$ ### Step 2: Forming the Final Equation We are given f'(1) = 2024$f'(1) = 2024$. Substituting this value: 2024 cdot frac1x = fracf'(x)f(x)$$2024 \cdot \frac{1}{x} = \frac{f'(x)}{f(x)}$$ Cross-multiplying yields: 2024 f(x) = x f'(x)$$2024 f(x) = x f'(x)$$ Rightarrow x f'(x) - 2024 f(x) = 0$$\Rightarrow x f'(x) - 2024 f(x) = 0$$ ### Pattern Recognition Functional equations of the form f(x/y) = f(x)/f(y)$f(x/y) = f(x)/f(y)$ inherently describe power functions (f(x) = x^k$f(x) = x^k$). Partial differentiation with respect to one variable quickly resolves the derivative form without limits. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Functions Class 12 Maths: Differential Equations

### 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