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 first-order linear differential equation fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$: - Integrating Factor (I.F.) = e^int P(x) \, dx$e^{\int P(x) \, dx}$ - Solution is y cdot textI.F. = int Q(x) cdot textI.F. \, dx + C$y \cdot \text{I.F.} = \int Q(x) \cdot \text{I.F.} \, dx + C$ ### Core Logic Let's simplify the coefficient of y$y$: 3tan^2 x + 3 = 3(tan^2 x + 1) = 3sec^2 x$$3\tan^2 x + 3 = 3(\tan^2 x + 1) = 3\sec^2 x$$ Thus, the equation is: fracdydx + 3(sec^2 x)y = sec^2 x$$\frac{dy}{dx} + 3(\sec^2 x)y = \sec^2 x$$ ### Step 1: Finding Integrating Factor and General Solution I.F. = e^int 3sec^2 x \, dx = e^3tan x$e^{\int 3\sec^2 x \, dx} = e^{3\tan x}$ The general solution is: y cdot e^3tan x = int sec^2 x cdot e^3tan x \, dx + C$$y \cdot e^{3\tan x} = \int \sec^2 x \cdot e^{3\tan x} \, dx + C$$ Substitute u = 3tan x implies du = 3sec^2 x \, dx$u = 3\tan x \implies du = 3\sec^2 x \, dx$: y cdot e^3tan x = frac13 int e^u \, du + C = frac13 e^3tan x + C$$y \cdot e^{3\tan x} = \frac{1}{3} \int e^u \, du + C = \frac{1}{3} e^{3\tan x} + C$$ ### Step 2: Solving for boundary conditions Given y(0) = frac13 + e^3$y(0) = \frac{1}{3} + e^3$: left(frac13 + e^3right) cdot e^0 = frac13 e^0 + C implies C = e^3$$\left(\frac{1}{3} + e^3\right) \cdot e^0 = \frac{1}{3} e^0 + C \implies C = e^3$$ Thus, the explicit function is: y = frac13 + e^3 - 3tan x$$y = \frac{1}{3} + e^{3 - 3\tan x}$$ Evaluating at x = fracpi4$x = \frac{\pi}{4}$: yleft(fracpi4right) = frac13 + e^3 - 3tan(pi/4) = frac13 + e^3-3 = frac13 + 1 = frac43$$y\left(\frac{\pi}{4}\right) = \frac{1}{3} + e^{3 - 3\tan(\pi/4)} = \frac{1}{3} + e^{3-3} = \frac{1}{3} + 1 = \frac{4}{3}$$ ### Pattern Recognition Recognizing that 3tan^2 x + 3 = 3sec^2 x$3\tan^2 x + 3 = 3\sec^2 x$ converts the system immediately into a classic linear differential equation where the coefficient of y$y$ is exactly the derivative of the exponent of I.F. This makes integration virtually instantaneous. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 11 Mathematics: Trigonometric Functions

### Related Formula Standard first-order linear differential equation form: fracdydx + P(x)y = Q(x)$$\frac{dy}{dx} + P(x)y = Q(x)$$ Integrating Factor: I.F. = e^int P(x)dx$$I.F. = e^{\int P(x)dx}$$ Solution layout: y cdot (I.F.) = int Q(x) cdot (I.F.) dx + C$$y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C$$ ### Core Logic Rearrange the given differential equation into standard linear form: x(x^2 + e^x)fracdydx + e^x(x - 2)y = x^3$$x(x^2 + e^x)\frac{dy}{dx} + e^x(x - 2)y = x^3$$ fracdydx + frace^x(x - 2)x(x^2 + e^x)y = fracx^2x^2 + e^x$$\frac{dy}{dx} + \frac{e^x(x - 2)}{x(x^2 + e^x)}y = \frac{x^2}{x^2 + e^x}$$ ### Step 1: Determine Integrating Factor We need to integrate P(x) = frace^x(x-2)x(x^2+e^x)$P(x) = \frac{e^x(x-2)}{x(x^2+e^x)}$: int frace^x(x-2)x(x^2+e^x) dx = int fracxe^x - 2e^xx(x^2+e^x) dx = int fracfracxe^x - 2e^xx^31 + frace^xx^2 dx$$\int \frac{e^x(x-2)}{x(x^2+e^x)} dx = \int \frac{xe^x - 2e^x}{x(x^2+e^x)} dx = \int \frac{\frac{xe^x - 2e^x}{x^3}}{1 + \frac{e^x}{x^2}} dx$$ Let t = 1 + frace^xx^2$t = 1 + \frac{e^x}{x^2}$. Then dt = fracx^2e^x - e^x(2x)x^4 dx = fracxe^x - 2e^xx^3 dx$dt = \frac{x^2e^x - e^x(2x)}{x^4} dx = \frac{xe^x - 2e^x}{x^3} dx$. Thus: int P(x)dx = int fracdtt = ln|t| = lnleft|1 + frace^xx^2right|$$\int P(x)dx = \int \frac{dt}{t} = \ln|t| = \ln\left|1 + \frac{e^x}{x^2}\right|$$ Therefore, the integrating factor is: I.F. = e^lnleft(1 + frace^xx^2right) = 1 + frace^xx^2 = fracx^2 + e^xx^2$$I.F. = e^{\ln\left(1 + \frac{e^x}{x^2}\right)} = 1 + \frac{e^x}{x^2} = \frac{x^2 + e^x}{x^2}$$ ### Step 2: Construct the General Solution Using the linear equation solution template: y cdot left(fracx^2 + e^xx^2right) = int left(fracx^2x^2 + e^xright) cdot left(fracx^2 + e^xx^2right) dx + C$$y \cdot \left(\frac{x^2 + e^x}{x^2}\right) = \int \left(\frac{x^2}{x^2 + e^x}\right) \cdot \left(\frac{x^2 + e^x}{x^2}\right) dx + C$$ y cdot left(1 + frace^xx^2right) = int 1 cdot dx + C$$y \cdot \left(1 + \frac{e^x}{x^2}\right) = \int 1 \cdot dx + C$$ y cdot left(1 + frace^xx^2right) = x + C$$y \cdot \left(1 + \frac{e^x}{x^2}\right) = x + C$$ ### Step 3: Evaluate Constant and Compute y(2) The curve passes through (1, 0)$(1, 0)$. Substitute x=1, y=0$x=1, y=0$: 0 cdot (1 + e) = 1 + C implies C = -1$$0 \cdot (1 + e) = 1 + C \implies C = -1$$ So the exact solution equation is: y cdot left(1 + frace^xx^2right) = x - 1 implies y = fracx - 11 + frace^xx^2$$y \cdot \left(1 + \frac{e^x}{x^2}\right) = x - 1 \implies y = \frac{x - 1}{1 + \frac{e^x}{x^2}}$$ To find y(2)$y(2)$, substitute x=2$x=2$: y(2) = frac2 - 11 + frace^22^2 = frac11 + frace^24 = frac44 + e^2$$y(2) = \frac{2 - 1}{1 + \frac{e^2}{2^2}} = \frac{1}{1 + \frac{e^2}{4}} = \frac{4}{4 + e^2}$$ ### Pattern Recognition Spotting the functional derivative framework inside P(x)$P(x)$ by dividing the numerator and denominator by x^3$x^3$ uncovers the standard form int fracf'(x)f(x)dx$\int \frac{f'(x)}{f(x)}dx$ cleanly, converting an otherwise intimidating integral into a basic natural log operation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integrals

### 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