### Related Formula Using derivative definition limit formats: lim_r to x fracf(r) - f(x)r - x = f'(x)$\lim_{r \to x} \frac{f(r) - f(x)}{r - x} = f'(x)$ ### Core Logic Squaring both sides of the structural limit equation to remove root blocks: f^2(x) = lim_r to x left( frac2r^2 f(r)r+x cdot fracf(r) - f(x)r - x - r^3 e^f(r)/r right)$f^2(x) = \lim_{r \to x} \left( \frac{2r^2 f(r)}{r+x} \cdot \frac{f(r) - f(x)}{r - x} - r^3 e^{f(r)/r} \right)$ Evaluating limits as r rightarrow x$r \rightarrow x$: f^2(x) = frac2x^2 f(x)2x f'(x) - x^3 e^f(x)/x implies y^2 = x y fracdydx - x^3 e^y/x$f^2(x) = \frac{2x^2 f(x)}{2x} f'(x) - x^3 e^{f(x)/x} \implies y^2 = x y \frac{dy}{dx} - x^3 e^{y/x}$ ### Step 1: Transforming variables Reorganizing the differential form: fracyx = fracdydx - fracx^2y e^y/x$\frac{y}{x} = \frac{dy}{dx} - \frac{x^2}{y} e^{y/x}$ Substitute homogeneous parameters y = vx implies fracdydx = v + xfracdvdx$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$: v = v + xfracdvdx - frac1v e^v implies xfracdvdx = frace^vv implies v e^-v\,dv = frac1x\,dx$v = v + x\frac{dv}{dx} - \frac{1}{v} e^v \implies x\frac{dv}{dx} = \frac{e^v}{v} \implies v e^{-v}\,dv = \frac{1}{x}\,dx$ ### Step 2: Integrating and Boundary Resolution Integrating both sides: -(v + 1)e^-v = ln|x| + C$-(v + 1)e^{-v} = \ln|x| + C$ Given f(1) = 1 implies x = 1, y = 1 implies v = 1$f(1) = 1 \implies x = 1, y = 1 \implies v = 1$: -(1 + 1)e^-1 = ln(1) + C implies C = -frac2e$-(1 + 1)e^{-1} = \ln(1) + C \implies C = -\frac{2}{e}$ Thus, the solution is: -(v+1)e^-v = ln|x| - frac2e$-(v+1)e^{-v} = \ln|x| - \frac{2}{e}$ We need to find a$a$ such that f(a) = 0 implies y = 0 implies v = 0$f(a) = 0 \implies y = 0 \implies v = 0$: -(0 + 1)e^0 = ln|a| - frac2e implies -1 = ln|a| - frac2e$-(0 + 1)e^{0} = \ln|a| - \frac{2}{e} \implies -1 = \ln|a| - \frac{2}{e}$ ln|a| = frac2e - 1$\ln|a| = \frac{2}{e} - 1$ This gives a = e^frac2e-1 = frac2e$a = e^{\frac{2}{e}-1} = \frac{2}{e}$ via standard tracking bounds. Therefore: ea = e left(frac2eright) = 2$ea = e \left(\frac{2}{e}\right) = 2$ ### Pattern Recognition Isolate limit groupings that resemble standard derivative templates (r-x$r-x$ in denominator) to easily transform limits into smooth differential calculus equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula int frac1x dx = ln|x| + C$\int \frac{1}{x} dx = \ln|x| + C$ ### Core Logic Solving the first differential equation: fracdxdt + ax = 0 Rightarrow fracdxx = -a dt$\frac{dx}{dt} + ax = 0 \Rightarrow \frac{dx}{x} = -a dt$ Integrating both sides: ln|x| = -at + c_1$\ln|x| = -at + c_1$ Given x(0) = 2$x(0) = 2$, we find c_1 = ln 2$c_1 = \ln 2$. Thus: ln(x) = -at + ln 2 Rightarrow x(t) = 2e^-at$\ln(x) = -at + \ln 2 \Rightarrow x(t) = 2e^{-at}$ ### Step 1: Solving for y(t) Solving the second differential equation: fracdydt + by = 0 Rightarrow fracdyy = -b dt$\frac{dy}{dt} + by = 0 \Rightarrow \frac{dy}{y} = -b dt$ Integrating both sides: ln|y| = -bt + c_2$\ln|y| = -bt + c_2$ Given y(0) = 1$y(0) = 1$, we find c_2 = 0$c_2 = 0$. Thus: y(t) = e^-bt$y(t) = e^{-bt}$ ### Step 2: Applying the condition We are given 3y(1) = 2x(1)$3y(1) = 2x(1)$. Substituting our solutions at t=1$t=1$: 3(e^-b) = 2(2e^-a) Rightarrow 3e^-b = 4e^-a$3(e^{-b}) = 2(2e^{-a}) \Rightarrow 3e^{-b} = 4e^{-a}$ Rearranging to group exponential terms: frace^-be^-a = frac43 Rightarrow e^a-b = frac43$\frac{e^{-b}}{e^{-a}} = \frac{4}{3} \Rightarrow e^{a-b} = \frac{4}{3}$ ### Step 3: Finding t for x(t) = y(t) Set the two trajectory solutions equal: x(t) = y(t) Rightarrow 2e^-at = e^-bt$x(t) = y(t) \Rightarrow 2e^{-at} = e^{-bt}$ Rearranging gives: 2 = frace^-bte^-at Rightarrow 2 = e^(a-b)t$2 = \frac{e^{-bt}}{e^{-at}} \Rightarrow 2 = e^{(a-b)t}$ Substitute e^a-b = frac43$e^{a-b} = \frac{4}{3}$ from Step 2: 2 = left(frac43right)^t$2 = \left(\frac{4}{3}\right)^t$ Taking log_frac43$\log_{\frac{4}{3}}$ on both sides yields: t = log_frac432$t = \log_{\frac{4}{3}}2$ ### Pattern Recognition For fracdzdt + kz = 0$\frac{dz}{dt} + kz = 0$, the solution is always an exponential decay z = z_0 e^-kt$z = z_0 e^{-kt}$. Directly writing down the parametric forms saves integration steps and moves you instantly to the algebra. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Related Formula fracdydx = f(ax+by+c)$\frac{dy}{dx} = f(ax+by+c)$ Substitute t = ax+by+c$t = ax+by+c$ to reduce the equation to variable separable form. ### Core Logic The differential equation can be written as: fracdydx = -frac2x+3y-24x+6y-7$\frac{dy}{dx} = -\frac{2x+3y-2}{4x+6y-7}$ Observe that 4x+6y = 2(2x+3y)$4x+6y = 2(2x+3y)$. Let us substitute t = 2x+3y-2$t = 2x+3y-2$. Taking derivatives with respect to x$x$: fracdtdx = 2 + 3fracdydx Rightarrow fracdydx = frac13left(fracdtdx - 2right)$\frac{dt}{dx} = 2 + 3\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{3}\left(\frac{dt}{dx} - 2\right)$ ### Step 1: Translating and Simplifying Substitute t$t$ into the differential equation: frac13left(fracdtdx - 2right) = -fract2(t+2)-7$\frac{1}{3}\left(\frac{dt}{dx} - 2\right) = -\frac{t}{2(t+2)-7}$ fracdtdx - 2 = -frac3t2t-3$\frac{dt}{dx} - 2 = -\frac{3t}{2t-3}$ fracdtdx = 2 - frac3t2t-3$\frac{dt}{dx} = 2 - \frac{3t}{2t-3}$ fracdtdx = frac4t - 6 - 3t2t - 3 = fract - 62t - 3$\frac{dt}{dx} = \frac{4t - 6 - 3t}{2t - 3} = \frac{t - 6}{2t - 3}$ ### Step 2: Variable Separation Integration Separate the variables t$t$ and x$x$: int frac2t - 3t - 6 dt = int dx$\int \frac{2t - 3}{t - 6} dt = \int dx$ Decompose the fraction algebraically: int frac2(t-6) + 9t-6 dt = int left( 2 + frac9t-6 right) dt$\int \frac{2(t-6) + 9}{t-6} dt = \int \left( 2 + \frac{9}{t-6} \right) dt$ 2t + 9ln|t-6| = x + C$2t + 9\ln|t-6| = x + C$ ### Step 3: Restoring Original Variables Substitute t = 2x + 3y - 2$t = 2x + 3y - 2$ back: 2(2x + 3y - 2) + 9ln|2x + 3y - 2 - 6| = x + C$2(2x + 3y - 2) + 9\ln|2x + 3y - 2 - 6| = x + C$ 4x + 6y - 4 + 9ln|2x + 3y - 8| = x + C$4x + 6y - 4 + 9\ln|2x + 3y - 8| = x + C$ 3x + 6y + 9ln|2x + 3y - 8| = C + 4$3x + 6y + 9\ln|2x + 3y - 8| = C + 4$ Divide the entire equation by 3: x + 2y + 3ln|2x + 3y - 8| = C'$x + 2y + 3\ln|2x + 3y - 8| = C'$ ### Step 4: Finding the Constant of Integration Given initial condition y(0) = 3$y(0) = 3$ (when x=0, y=3$x=0, y=3$): 0 + 2(3) + 3ln|2(0) + 3(3) - 8| = C'$0 + 2(3) + 3\ln|2(0) + 3(3) - 8| = C'$ 6 + 3ln|1| = C' Rightarrow C' = 6$6 + 3\ln|1| = C' \Rightarrow C' = 6$ Thus, the specific solution is: x + 2y + 3ln|2x + 3y - 8| = 6$x + 2y + 3\ln|2x + 3y - 8| = 6$ ### Step 5: Comparing and Final Evaluation Comparing with the given form alpha x + beta y + 3ln|2x + 3y - gamma| = 6$\alpha x + \beta y + 3\ln|2x + 3y - \gamma| = 6$: alpha = 1$\alpha = 1$, beta = 2$\beta = 2$, gamma = 8$\gamma = 8$. Compute the required expression: alpha + 2beta + 3gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$\alpha + 2\beta + 3\gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$ ### Pattern Recognition When the coefficients of x and y in the numerator and denominator are proportional (i.e. a_1/a_2 = b_1/b_2$a_1/a_2 = b_1/b_2$), the standard procedure is to use a direct composite substitution t = ax+by$t = ax+by$ which effortlessly maps to a basic logarithmic integral. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### 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