### 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

### 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