Let y = y(x)$y = y(x)$ be the solution of the differential equation (x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)cos x$(x^2 + 1)y' - 2xy = (x^4 + 2x^2 + 1)\cos x$, y(0) = 1$y(0) = 1$. Then int_-3^3 y(x) dx$\int_{-3}^{3} y(x) dx$ is:

### 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 textLeibniz Rule for Differentiation under Integral Sign: fracddx int_u(x)^v(x) f(t) dt = f(v(x)) v'(x) - f(u(x)) u'(x)$$\text{Leibniz Rule for Differentiation under Integral Sign: } \frac{d}{dx} \int_{u(x)}^{v(x)} f(t) dt = f(v(x)) v'(x) - f(u(x)) u'(x)$$ textStandard Linear Differential Equation: fracdydx + P(x) y = Q(x)$$\text{Standard Linear Differential Equation: } \frac{dy}{dx} + P(x) y = Q(x)$$ ### Core Logic Differentiating both sides with respect to x$x$ eliminates the definite integral and leads to a first-order linear differential equation. ### Step 1: Differentiate both sides Apply differentiation with respect to x$x$ using the Leibniz Rule on the left side, and product rule on the right side: 10 f(x) = 5 f(x) + 5x f'(x) - 5x^4$$10 f(x) = 5 f(x) + 5x f'(x) - 5x^4$$ 5 f(x) + 5x^4 = 5x f'(x) implies f(x) + x^4 = x f'(x)$$5 f(x) + 5x^4 = 5x f'(x) \implies f(x) + x^4 = x f'(x)$$ Letting y = f(x)$y = f(x)$, we rewrite it as: fracdydx - fracyx = x^3$$\frac{dy}{dx} - \frac{y}{x} = x^3$$ ### Step 2: Solve the Linear Differential Equation The integrating factor (I.F.) is: textI.F. = e^int -frac1x dx = e^-ln x = frac1x$$\text{I.F.} = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x}$$ Multiply by the I.F. and integrate: y cdot frac1x = int x^3 cdot frac1x dx = int x^2 dx = fracx^33 + C$$y \cdot \frac{1}{x} = \int x^3 \cdot \frac{1}{x} dx = \int x^2 dx = \frac{x^3}{3} + C$$ Thus, the general solution is: f(x) = fracx^43 + C x$$f(x) = \frac{x^4}{3} + C x$$ ### Step 3: Apply Boundary Conditions Substitute x = 1$x = 1$ into the original integral equation: 10 int_1^1 f(t) dt = 5(1) f(1) - 1^5 - 9$$10 \int_{1}^{1} f(t) dt = 5(1) f(1) - 1^5 - 9$$ 0 = 5 f(1) - 10 implies f(1) = 2$$0 = 5 f(1) - 10 \implies f(1) = 2$$ Now, substitute x = 1$x = 1$ and f(1) = 2$f(1) = 2$ into our general solution to find C$C$: 2 = frac13 + C implies C = frac53$$2 = \frac{1}{3} + C \implies C = \frac{5}{3}$$ Therefore, the complete function is: f(x) = fracx^43 + frac5x3$$f(x) = \frac{x^4}{3} + \frac{5x}{3}$$ ### Step 4: Compute f(3) Evaluate the function at x = 3$x = 3$: f(3) = frac3^43 + frac5(3)3 = 27 + 5 = 32$$f(3) = \frac{3^4}{3} + \frac{5(3)}{3} = 27 + 5 = 32$$ ### Pattern Recognition Whenever a definite integral is defined from a constant to the variable x$x$ within an equation, differentiating immediately reduces it to a differential equation. Finding the value of f(a)$f(a)$ at the lower bound is a standard method to get the constant of integration. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula textLinear Differential equation form: fracdydx + P(x) y = Q(x)$$\text{Linear Differential equation form: } \frac{dy}{dx} + P(x) y = Q(x)$$ textIntegrating Factor: I.F. = e^int P(x) dx$$\text{Integrating Factor: } I.F. = e^{\int P(x) dx}$$ textGeneral solution: y cdot I.F. = int Q(x) cdot I.F. \, dx + C$$\text{General solution: } y \cdot I.F. = \int Q(x) \cdot I.F. \, dx + C$$ ### Core Logic This is a first-order linear differential equation. We calculate the Integrating Factor first to write down the integral solution. ### Step 1: Find the Integrating Factor (I.F.) Here, P(x) = 2sec^2 x$P(x) = 2\sec^2 x$ and Q(x) = 2sec^2 x + 3tan x cdot sec^2 x$Q(x) = 2\sec^2 x + 3\tan x \cdot \sec^2 x$: textI.F. = e^int 2sec^2 x dx = e^2tan x$$\text{I.F.} = e^{\int 2\sec^2 x dx} = e^{2\tan x}$$ ### Step 2: Obtain the General Solution Multiply both sides by the integrating factor: y cdot e^2tan x = int e^2tan x left( 2sec^2 x + 3tan x cdot sec^2 x right) dx$$y \cdot e^{2\tan x} = \int e^{2\tan x} \left( 2\sec^2 x + 3\tan x \cdot \sec^2 x \right) dx$$ Let t = tan x implies dt = sec^2 x dx$t = \tan x \implies dt = \sec^2 x dx$. The integral becomes: int e^2t (2 + 3t) dt = int 2 e^2t dt + 3 int t e^2t dt$$\int e^{2t} (2 + 3t) dt = \int 2 e^{2t} dt + 3 \int t e^{2t} dt$$ Applying integration by parts for the second term: 3 int t e^2t dt = 3 left[ fract e^2t2 - int frace^2t2 dt right] = frac3t e^2t2 - frac3e^2t4$$3 \int t e^{2t} dt = 3 \left[ \frac{t e^{2t}}{2} - \int \frac{e^{2t}}{2} dt \right] = \frac{3t e^{2t}}{2} - \frac{3e^{2t}}{4}$$ Summing all parts: y cdot e^2tan x = e^2t + frac3t e^2t2 - frac3e^2t4 + C = e^2tan x left[ 1 + frac3tan x2 - frac34 right] + C$$y \cdot e^{2\tan x} = e^{2t} + \frac{3t e^{2t}}{2} - \frac{3e^{2t}}{4} + C = e^{2\tan x} \left[ 1 + \frac{3\tan x}{2} - \frac{3}{4} \right] + C$$ y = frac3tan x2 + frac14 + C e^-2tan x$$y = \frac{3\tan x}{2} + \frac{1}{4} + C e^{-2\tan x}$$ ### Step 3: Apply the boundary conditions Using the initial boundary condition y(0) = frac54$y(0) = \frac{5}{4}$: frac54 = frac3(0)2 + frac14 + C e^0 implies C = 1$$\frac{5}{4} = \frac{3(0)}{2} + \frac{1}{4} + C e^{0} \implies C = 1$$ Thus, the complete function is: y(x) = frac3tan x2 + frac14 + e^-2tan x$$y(x) = \frac{3\tan x}{2} + \frac{1}{4} + e^{-2\tan x}$$ ### Step 4: Compute the final value Evaluate the function at x = fracpi4$x = \frac{\pi}{4}$: yleft(fracpi4right) = frac3(1)2 + frac14 + e^-2 = frac74 + e^-2$$y\left(\frac{\pi}{4}\right) = \frac{3(1)}{2} + \frac{1}{4} + e^{-2} = \frac{7}{4} + e^{-2}$$ Now, calculate the requested value: 12 left( yleft(fracpi4right) - e^-2 right) = 12 left( frac74 right) = 21$$12 \left( y\left(\frac{\pi}{4}\right) - e^{-2} \right) = 12 \left( \frac{7}{4} \right) = 21$$ ### Pattern Recognition Integration by parts substitution: When integrating terms of the form int e^at P(t) dt$\int e^{at} P(t) dt$ (where P(t)$P(t)$ is a polynomial), substituting the polynomial variable directly simplifies the exponential integration factors cleanly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula Standard variable separable integration form: int frac1sqrty^2 + a^2 \, mathrmdy = lnleft|y + sqrty^2 + a^2right| + C$$\int \frac{1}{\sqrt{y^2 + a^2}} \, \mathrm{d}y = \ln\left|y + \sqrt{y^2 + a^2}\right| + C$$ ### Core Logic The original paper solution states the structure equation setup as f''(x) = f(x)$f''(x) = f(x)$. Multiply by f'(x)$f'(x)$ on both sides to transform it into a integrable derivative form. ### Step 1: Integrate the derivative identity f'(x) cdot f''(x) = f'(x) cdot f(x)$$f'(x) \cdot f''(x) = f'(x) \cdot f(x)$$ Integrate both sides with respect to x$x$: frac(f'(x))^22 = frac(f(x))^22 + C implies (f'(x))^2 = (f(x))^2 + C'$$\frac{(f'(x))^2}{2} = \frac{(f(x))^2}{2} + C \implies (f'(x))^2 = (f(x))^2 + C'$$ ### Step 2: Find the constant of integration Use initial conditions f(0) = 0$f(0) = 0$ and f'(0) = 3$f'(0) = 3$: 3^2 = 0^2 + C' implies C' = 9$$3^2 = 0^2 + C' \implies C' = 9$$ Thus, (f'(x))^2 = (f(x))^2 + 9$(f'(x))^2 = (f(x))^2 + 9$. Given f'(x) ge 0$f'(x) \ge 0$: f'(x) = sqrt(f(x))^2 + 9$$f'(x) = \sqrt{(f(x))^2 + 9}$$ ### Step 3: Variable Separation and Solution Form Let y = f(x) implies fracmathrmdymathrmdx = sqrty^2 + 9$y = f(x) \implies \frac{\mathrm{d}y}{\mathrm{d}x} = \sqrt{y^2 + 9}$: int fracmathrmdysqrty^2 + 9 = int mathrmdx implies lnleft|y + sqrty^2 + 9right| = x + C_2$$\int \frac{\mathrm{d}y}{\sqrt{y^2 + 9}} = \int \mathrm{d}x \implies \ln\left|y + \sqrt{y^2 + 9}\right| = x + C_2$$ Substitute initial condition x=0, y=0$x=0, y=0$: ln|0 + sqrt9| = 0 + C_2 implies C_2 = ln 3$$\ln|0 + \sqrt{9}| = 0 + C_2 \implies C_2 = \ln 3$$ Therefore, lnleft|y + sqrty^2 + 9right| = x + ln 3 implies y + sqrty^2 + 9 = 3e^x$\ln\left|y + \sqrt{y^2 + 9}\right| = x + \ln 3 \implies y + \sqrt{y^2 + 9} = 3e^x$. ### Step 4: Compute targeted value We need to evaluate at x = ln 3$x = \ln 3$: y + sqrty^2 + 9 = 3e^ln 3 = 3(3) = 9$$y + \sqrt{y^2 + 9} = 3e^{\ln 3} = 3(3) = 9$$ sqrty^2 + 9 = 9 - y$$\sqrt{y^2 + 9} = 9 - y$$ Square both sides: y^2 + 9 = 81 - 18y + y^2 implies 18y = 72 implies y = 4$$y^2 + 9 = 81 - 18y + y^2 \implies 18y = 72 \implies y = 4$$ Thus, f(ln 3) = 4$f(\ln 3) = 4$. Multiply by 9$9$: 9 f(ln 3) = 9(4) = 36$$9 f(\ln 3) = 9(4) = 36$$ ### Pattern Recognition Multiplying a second derivative by the first derivative (f'f''$f'f''$) is a classic trick to convert a second-order linear differential equation into a first-order separable layout, opening a clear path to the solution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Differential Calculus