Let the solution curve of the differential equation x \, dy - y \, dx = sqrtx^2 + y^2 \, dx, \, x > 0, \, y(1) = 0,$x \, dy - y \, dx = \sqrt{x^{2} + y^{2}} \, dx, \, x > 0, \, y(1) = 0,$ be y = y(x)$y = y(x)$. Then y(3)$y(3)$ is equal to

### Related Formula For a linear differential equation of the form fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$: textIntegrating Factor (IF) = e^int P(x)dx$$\text{Integrating Factor (IF)} = e^{\int P(x)dx}$$ textSolution is y cdot textIF = int (Q(x) cdot textIF) dx + C$$\text{Solution is } y \cdot \text{IF} = \int (Q(x) \cdot \text{IF}) dx + C$$ ### Core Logic Rearrange the given differential equation to standard linear form: (1 + x^2)dy = (tan^-1x - y)dx$(1 + x^2)dy = (\tan^{-1}x - y)dx$ fracdydx + fracyx^2 + 1 = fractan^-1xx^2 + 1$$\frac{dy}{dx} + \frac{y}{x^2 + 1} = \frac{\tan^{-1}x}{x^2 + 1}$$ ### Step 1: Find the Integrating Factor P(x) = frac1x^2 + 1$$P(x) = \frac{1}{x^2 + 1}$$ textIF = e^int frac1x^2 + 1 dx = e^tan^-1x$$\text{IF} = e^{\int \frac{1}{x^2 + 1} dx} = e^{\tan^{-1}x}$$ ### Step 2: Solve the Integral y cdot e^tan^-1x = int e^tan^-1x cdot fractan^-1x1 + x^2 dx$$y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{\tan^{-1}x}{1 + x^2} dx$$ Let t = tan^-1x$t = \tan^{-1}x$, then dt = frac11 + x^2 dx$dt = \frac{1}{1 + x^2} dx$. The integral becomes int t e^t dt$\int t e^t dt$. Using integration by parts: int t e^t dt = t e^t - e^t + C$$\int t e^t dt = t e^t - e^t + C$$ Substituting back t = tan^-1x$t = \tan^{-1}x$: y cdot e^tan^-1x = tan^-1x(e^tan^-1x) - e^tan^-1x + C$$y \cdot e^{\tan^{-1}x} = \tan^{-1}x(e^{\tan^{-1}x}) - e^{\tan^{-1}x} + C$$ ### Step 3: Apply Boundary Condition Given y(0) = 1$y(0) = 1$: 1 cdot e^0 = 0 cdot e^0 - e^0 + C$$1 \cdot e^0 = 0 \cdot e^0 - e^0 + C$$ 1 = 0 - 1 + C Rightarrow C = 2$$1 = 0 - 1 + C \Rightarrow C = 2$$ ### Step 4: Evaluate at x = 1 Equation of curve: y = tan^-1x - 1 + 2e^-tan^-1x$$y = \tan^{-1}x - 1 + 2e^{-\tan^{-1}x}$$ Evaluate at x = 1$x = 1$: y(1) = tan^-1(1) - 1 + 2e^-tan^-1(1)$$y(1) = \tan^{-1}(1) - 1 + 2e^{-\tan^{-1}(1)}$$ y(1) = fracpi4 - 1 + 2e^-pi/4$$y(1) = \frac{\pi}{4} - 1 + 2e^{-\pi/4}$$ y(1) = frac2e^pi/4 + fracpi4 - 1$$y(1) = \frac{2}{e^{\pi/4}} + \frac{\pi}{4} - 1$$ ### Pattern Recognition A classic LDE integration trap: int e^f(x) f(x) f'(x) dx$\int e^{f(x)} f(x) f'(x) dx$ resolves trivially with the substitution u = f(x)$u = f(x)$ turning it into int u e^u du$\int u e^u du$, which always evaluates to e^u(u - 1) + C$e^u(u - 1) + C$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Integrals

### Related Formula For a quadratic equation Am^2 + Bm + C = 0$Am^2 + Bm + C = 0$ having equal roots, Discriminant D = 0 Rightarrow B^2 - 4AC = 0$D = 0 \Rightarrow B^2 - 4AC = 0$. ### Core Logic Given quadratic equation in m$m$: f(x)m^2 - 2f'(x)m + f''(x) = 0$f(x)m^2 - 2f'(x)m + f''(x) = 0$ has equal roots. D = 0 Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0$$D = 0 \Rightarrow (-2f'(x))^2 - 4(f(x))(f''(x)) = 0$$ 4(f'(x))^2 = 4f(x)f''(x) Rightarrow (f'(x))^2 = f(x)f''(x)$$4(f'(x))^2 = 4f(x)f''(x) \Rightarrow (f'(x))^2 = f(x)f''(x)$$ ### Step 1: Solve the Differential Equation Rewrite the DE: fracf''(x)f'(x) = fracf'(x)f(x)$\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$ Integrate both sides: int fracf''(x)f'(x) dx = int fracf'(x)f(x) dx$$\int \frac{f''(x)}{f'(x)} dx = \int \frac{f'(x)}{f(x)} dx$$ ln|f'(x)| = ln|f(x)| + ln|c| Rightarrow f'(x) = c cdot f(x)$$\ln|f'(x)| = \ln|f(x)| + \ln|c| \Rightarrow f'(x) = c \cdot f(x)$$ Using given f(0) = 1$f(0) = 1$ and f'(0) = 2$f'(0) = 2$: f'(0) = c cdot f(0) Rightarrow 2 = c(1) Rightarrow c = 2$$f'(0) = c \cdot f(0) \Rightarrow 2 = c(1) \Rightarrow c = 2$$ Now we have f'(x) = 2f(x) Rightarrow fracf'(x)f(x) = 2$f'(x) = 2f(x) \Rightarrow \frac{f'(x)}{f(x)} = 2$. Integrate again: ln|f(x)| = 2x + d$$\ln|f(x)| = 2x + d$$ Use f(0) = 1 Rightarrow ln(1) = 0 + d Rightarrow d = 0$f(0) = 1 \Rightarrow \ln(1) = 0 + d \Rightarrow d = 0$. So, ln f(x) = 2x Rightarrow f(x) = e^2x$\ln f(x) = 2x \Rightarrow f(x) = e^{2x}$. ### Step 2: Investigate increasing interval Let g(x) = f(ln x - x) = e^2(ln x - x)$g(x) = f(\ln x - x) = e^{2(\ln x - x)}$. For g(x)$g(x)$ to be increasing, g'(x) geq 0$g'(x) \geq 0$. g'(x) = 2e^2(ln x - x) cdot fracddx(ln x - x)$$g'(x) = 2e^{2(\ln x - x)} \cdot \frac{d}{dx}(\ln x - x)$$ g'(x) = 2e^2(ln x - x) left(frac1x - 1right)$$g'(x) = 2e^{2(\ln x - x)} \left(\frac{1}{x} - 1\right)$$ Since exponential is always positive, g'(x) geq 0 Rightarrow frac1x - 1 geq 0$g'(x) \geq 0 \Rightarrow \frac{1}{x} - 1 \geq 0$. frac1 - xx geq 0$$\frac{1 - x}{x} \geq 0$$ The critical points are x=0, x=1$x=0, x=1$. Based on domain of ln x$\ln x$, x > 0$x > 0$. Sign scheme yields positive derivative in x in (0, 1]$x \in (0, 1]$. Therefore, (alpha, beta) = (0, 1) Rightarrow alpha = 0, beta = 1$(\alpha, \beta) = (0, 1) \Rightarrow \alpha = 0, \beta = 1$. ### Step 3: Final Output alpha + beta = 0 + 1 = 1$$\alpha + \beta = 0 + 1 = 1$$ ### Pattern Recognition The relation (f')^2 = f cdot f''$(f')^2 = f \cdot f''$ is a classical indicator of exponential functions (f = Ce^kx$f = Ce^{kx}$). Solving via double logarithmic integration collapses the differential equation almost instantaneously. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Applications of Derivatives

### Related Formula textStandard form LDE: fracdydx + P(x)y = Q(x)$$\text{Standard form LDE: } \frac{dy}{dx} + P(x)y = Q(x)$$ textIntegrating Factor (I.F.) = e^int P(x)dx$$\text{Integrating Factor (I.F.) } = e^{\int P(x)dx}$$ textSolution is y(I.F.) = int Q(x)(I.F.)dx + C$$\text{Solution is } y(I.F.) = \int Q(x)(I.F.)dx + C$$ ### Core Logic Multiply the entire differential equation by cos x$\cos x$ to normalize the fracdydx$\frac{dy}{dx}$ term: fracdydx - (2cos x)y = 2cos x + 3sin xcos x$$\frac{dy}{dx} - (2\cos x)y = 2\cos x + 3\sin x\cos x$$ This is a standard first-order linear differential equation with P(x) = -2cos x$P(x) = -2\cos x$. ### Step 1: Integrating Factor textI.F. = e^int -2cos x \, dx = e^-2sin x$$\text{I.F.} = e^{\int -2\cos x \, dx} = e^{-2\sin x}$$ ### Step 2: General Solution y cdot e^-2sin x = int e^-2sin x (2cos x + 3sin xcos x) \, dx$$y \cdot e^{-2\sin x} = \int e^{-2\sin x} (2\cos x + 3\sin x\cos x) \, dx$$ Let u = sin x implies du = cos x \, dx$u = \sin x \implies du = \cos x \, dx$. The integral becomes: int e^-2u (2 + 3u) \, du$$\int e^{-2u} (2 + 3u) \, du$$ Using integration by parts: = (2+3u)left(frace^-2u-2right) - int 3 left(frace^-2u-2right) \, du$$= (2+3u)\left(\frac{e^{-2u}}{-2}\right) - \int 3 \left(\frac{e^{-2u}}{-2}\right) \, du$$ = -frac2+3u2 e^-2u - frac34 e^-2u$$= -\frac{2+3u}{2} e^{-2u} - \frac{3}{4} e^{-2u}$$ = e^-2u left(-frac2+3u2 - frac34right) = e^-2u left(-frac4+6u+34right) = e^-2u left(-frac6u+74right)$$= e^{-2u} \left(-\frac{2+3u}{2} - \frac{3}{4}\right) = e^{-2u} \left(-\frac{4+6u+3}{4}\right) = e^{-2u} \left(-\frac{6u+7}{4}\right)$$ Re-substitute u = sin x$u = \sin x$: y e^-2sin x = e^-2sin x left(-frac32sin x - frac74right) + C$$y e^{-2\sin x} = e^{-2\sin x} \left(-\frac{3}{2}\sin x - \frac{7}{4}\right) + C$$ y = -frac32sin x - frac74 + C e^2sin x$$y = -\frac{3}{2}\sin x - \frac{7}{4} + C e^{2\sin x}$$ ### Step 3: Apply Boundary Conditions Given y(0) = -frac74$y(0) = -\frac{7}{4}$: -frac74 = -frac32(0) - frac74 + C e^0 implies C = 0$$-\frac{7}{4} = -\frac{3}{2}(0) - \frac{7}{4} + C e^0 \implies C = 0$$ Thus, y(x) = -frac32sin x - frac74$y(x) = -\frac{3}{2}\sin x - \frac{7}{4}$. ### Step 4: Find Final Value Calculate yleft(fracpi6right)$y\left(\frac{\pi}{6}\right)$: yleft(fracpi6right) = -frac32sinleft(fracpi6right) - frac74 = -frac32left(frac12right) - frac74$$y\left(\frac{\pi}{6}\right) = -\frac{3}{2}\sin\left(\frac{\pi}{6}\right) - \frac{7}{4} = -\frac{3}{2}\left(\frac{1}{2}\right) - \frac{7}{4}$$ = -frac34 - frac74 = -frac104 = -frac52$$= -\frac{3}{4} - \frac{7}{4} = -\frac{10}{4} = -\frac{5}{2}$$ ### Pattern Recognition When integrating e^ax(f(x))$e^{ax}(f(x))$, use DI method or standard Integration by Parts mapping. For int e^-2u(3u+2)$\int e^{-2u}(3u+2)$, always extract polynomial as u$u$, exponential as dv$dv$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Related Formula textLeibniz's Rule: fracddx int_a^x f(t) dt = f(x)$$\text{Leibniz's Rule: } \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x)$$ textLinear DE form: fracdydx + P(x)y = Q(x)$$\text{Linear DE form: } \frac{dy}{dx} + P(x)y = Q(x)$$ ### Core Logic Given 6int_1^xf(t)dt=3xf(x)+x^3-4$6\int_{1}^{x}f(t)dt=3xf(x)+x^{3}-4$. Differentiating both sides with respect to x$x$: 6 f(x) = 3(1 cdot f(x) + x f'(x)) + 3x^2$$6 f(x) = 3(1 \cdot f(x) + x f'(x)) + 3x^2$$ 6 f(x) = 3f(x) + 3x f'(x) + 3x^2$$6 f(x) = 3f(x) + 3x f'(x) + 3x^2$$ 3f(x) = 3x f'(x) + 3x^2$$3f(x) = 3x f'(x) + 3x^2$$ ### Step 1: Solving the Differential Equation Let y = f(x)$y = f(x)$. The equation simplifies to: x fracdydx - y = -x^2$$x \frac{dy}{dx} - y = -x^2$$ Divide the entire equation by x^2$x^2$ to recognize the exact differential form: fracx fracdydx - yx^2 = -1$$\frac{x \frac{dy}{dx} - y}{x^2} = -1$$ fracddx left( fracyx right) = -1$$\frac{d}{dx} \left( \frac{y}{x} \right) = -1$$ Integrate both sides: fracyx = -x + C$$\frac{y}{x} = -x + C$$ y = -x^2 + Cx implies f(x) = -x^2 + Cx$$y = -x^2 + Cx \implies f(x) = -x^2 + Cx$$ ### Step 2: Evaluating the Constant To find C$C$, we plug x=1$x=1$ into the original integral equation: 6int_1^1f(t)dt = 3(1)f(1) + 1^3 - 4$$6\int_{1}^{1}f(t)dt = 3(1)f(1) + 1^3 - 4$$ 0 = 3f(1) - 3 implies f(1) = 1$$0 = 3f(1) - 3 \implies f(1) = 1$$ Using f(x) = -x^2 + Cx$f(x) = -x^2 + Cx$: f(1) = -1 + C = 1 implies C = 2$$f(1) = -1 + C = 1 \implies C = 2$$ Thus, f(x) = -x^2 + 2x$f(x) = -x^2 + 2x$. ### Step 3: Final Calculation We need to find f(2) - f(3)$f(2) - f(3)$: f(2) = -(2)^2 + 2(2) = -4 + 4 = 0$$f(2) = -(2)^2 + 2(2) = -4 + 4 = 0$$ f(3) = -(3)^2 + 2(3) = -9 + 6 = -3$$f(3) = -(3)^2 + 2(3) = -9 + 6 = -3$$ f(2) - f(3) = 0 - (-3) = 3$$f(2) - f(3) = 0 - (-3) = 3$$ ### Pattern Recognition An integral equation involving a generic function f(t)$f(t)$ within variable limits signals the immediate use of Leibniz's rule. Differentiating leads cleanly to a linear differential equation. Re-substituting the lower limit (x=1$x=1$) securely yields the boundary condition f(1)$f(1)$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations Class 12 Maths: Applications of Integrals