Let y = y(x)$y = y(x)$ be the solution of the differential equation fracdydx + 3(tan^2 x)y + 3y = sec^2 x, y(0) = frac13 + e^3$\frac{dy}{dx} + 3(\tan^2 x)y + 3y = \sec^2 x, y(0) = \frac{1}{3} + e^3$. Then yleft(fracpi4 ight)$y\left(\frac{pi}{4} ight)$ is equal to

### Related Formula For a linear differential equation fracdydx + P(x)y = Q(x)$\frac{dy}{dx} + P(x)y = Q(x)$: IF = e^int P(x) dx$$IF = e^{\int P(x) dx}$$ Solution: y cdot IF = int Q(x) cdot IF \, dx + C$y \cdot IF = \int Q(x) \cdot IF \, dx + C$ ### Core Logic Rearrange the given differential equation into standard linear form: (1 - x^2) fracdydx = xy + (x^3 + 2)sqrt3(1 - x^2)$$(1 - x^2) \frac{dy}{dx} = xy + (x^3 + 2)\sqrt{3(1 - x^2)}$$ fracdydx - fracx1 - x^2 y = frac(x^3 + 2)sqrt3(1 - x^2)1 - x^2$$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{(x^3 + 2)\sqrt{3(1 - x^2)}}{1 - x^2}$$ fracdydx - fracx1 - x^2 y = fracsqrt3(x^3 + 2)sqrt1 - x^2$$\frac{dy}{dx} - \frac{x}{1 - x^2} y = \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}}$$ Identify P(x) = -fracx1 - x^2$P(x) = -\frac{x}{1 - x^2}$. ### Step 1: Finding Integrating Factor (IF) IF = e^int -fracx1 - x^2 dx$$IF = e^{\int -\frac{x}{1 - x^2} dx}$$ Let 1 - x^2 = t Rightarrow -2x dx = dt Rightarrow -x dx = fracdt2$1 - x^2 = t \Rightarrow -2x dx = dt \Rightarrow -x dx = \frac{dt}{2}$. IF = e^frac12 int frac1t dt = e^frac12 ln t = e^ln sqrtt = sqrt1 - x^2$$IF = e^{\frac{1}{2} \int \frac{1}{t} dt} = e^{\frac{1}{2} \ln t} = e^{\ln \sqrt{t}} = \sqrt{1 - x^2}$$ ### Step 2: General Solution The solution is given by: y sqrt1 - x^2 = int left( fracsqrt3(x^3 + 2)sqrt1 - x^2 right) sqrt1 - x^2 \, dx + C$$y \sqrt{1 - x^2} = \int \left( \frac{\sqrt{3}(x^3 + 2)}{\sqrt{1 - x^2}} \right) \sqrt{1 - x^2} \, dx + C$$ y sqrt1 - x^2 = sqrt3 int (x^3 + 2) \, dx + C$$y \sqrt{1 - x^2} = \sqrt{3} \int (x^3 + 2) \, dx + C$$ y sqrt1 - x^2 = sqrt3 left( fracx^44 + 2x right) + C$$y \sqrt{1 - x^2} = \sqrt{3} \left( \frac{x^4}{4} + 2x \right) + C$$ ### Step 3: Finding Constant C Using y(0) = 0$y(0) = 0$: 0 cdot 1 = sqrt3(0 + 0) + C Rightarrow C = 0$$0 \cdot 1 = \sqrt{3}(0 + 0) + C \Rightarrow C = 0$$ So, y(x) = fracsqrt3sqrt1 - x^2 left( fracx^44 + 2x right)$y(x) = \frac{\sqrt{3}}{\sqrt{1 - x^2}} \left( \frac{x^4}{4} + 2x \right)$. ### Step 4: Evaluating required point Substitute x = 1/2$x = 1/2$: yleft(frac12right) = fracsqrt3sqrt1 - 1/4 left( frac1/164 + 2left(frac12right) right)$$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{\sqrt{1 - 1/4}} \left( \frac{1/16}{4} + 2\left(\frac{1}{2}\right) \right)$$ Wait, frac(1/2)^44 = frac1/164 = frac164$\frac{(1/2)^4}{4} = \frac{1/16}{4} = \frac{1}{64}$. Let's recheck the expression: yleft(frac12right) = fracsqrt3sqrt3/4 left( frac164 + 1 right)$$y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{\sqrt{3/4}} \left( \frac{1}{64} + 1 \right)$$ = fracsqrt3sqrt3/2 left( frac6564 right) = 2 times frac6564 = frac6532$$= \frac{\sqrt{3}}{\sqrt{3}/2} \left( \frac{65}{64} \right) = 2 \times \frac{65}{64} = \frac{65}{32}$$ Here, m = 65$m = 65$ and n = 32$n = 32$. They are co-prime. Thus, m + n = 65 + 32 = 97$m + n = 65 + 32 = 97$. ### Pattern Recognition Whenever roots matching the integration denominator appear on the RHS of a linear differential setup, it is a high-confidence signal that the integrating factor cleanly annihilates the fractional root component during the solution stage. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Related Formula int fracdTT-T_s = -K int dt implies ln|T-T_s| = -Kt + C$$\int \frac{dT}{T-T_s} = -K \int dt \implies \ln|T-T_s| = -Kt + C$$ ### Core Logic Given fracdTdt = -K(T-80)$\frac{dT}{dt} = -K(T-80)$. Integrating from t=0$t=0$ to t$t$: int_160^T fracdTT-80 = -K int_0^t dt$$\int_{160}^{T} \frac{dT}{T-80} = -K \int_0^t dt$$ [ln|T-80|]_160^T = -Kt$$[\ln|T-80|]_{160}^T = -Kt$$ lnleft(fracT-8080right) = -Kt implies T(t) = 80 + 80e^-Kt$$\ln\left(\frac{T-80}{80}\right) = -Kt \implies T(t) = 80 + 80e^{-Kt}$$ Given T(15) = 120$T(15) = 120$: 120 = 80 + 80e^-15K implies 40 = 80e^-15K implies e^-15K = frac12$$120 = 80 + 80e^{-15K} \implies 40 = 80e^{-15K} \implies e^{-15K} = \frac{1}{2}$$ To find T(45)$T(45)$: T(45) = 80 + 80e^-45K = 80 + 80(e^-15K)^3$$T(45) = 80 + 80e^{-45K} = 80 + 80(e^{-15K})^3$$ = 80 + 80left(frac12right)^3$$= 80 + 80\left(\frac{1}{2}\right)^3$$ = 80 + 80left(frac18right) = 90$$= 80 + 80\left(\frac{1}{8}\right) = 90$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Related Formula textIntegrating factor for fracdudy + P(y)u = Q(y) text is I.F. = e^int P(y)dy$$\text{Integrating factor for } \frac{du}{dy} + P(y)u = Q(y) \text{ is } I.F. = e^{\int P(y)dy}$$ ### Core Logic Given DE: sec^2 x fracdxdy + e^2y tan^2 x + tan x = 0$\sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0$ Substitute t = tan x implies fracdtdy = sec^2 x fracdxdy$t = \tan x \implies \frac{dt}{dy} = \sec^2 x \frac{dx}{dy}$. The DE becomes: fracdtdy + t = -t^2 e^2y$$\frac{dt}{dy} + t = -t^2 e^{2y}$$ This is a Bernoulli equation in t$t$. Divide by t^2$t^2$: frac1t^2fracdtdy + frac1t = -e^2y$$\frac{1}{t^2}\frac{dt}{dy} + \frac{1}{t} = -e^{2y}$$ Substitute u = frac1t implies fracdudy = -frac1t^2fracdtdy$u = \frac{1}{t} \implies \frac{du}{dy} = -\frac{1}{t^2}\frac{dt}{dy}$. -fracdudy + u = -e^2y implies fracdudy - u = e^2y$$-\frac{du}{dy} + u = -e^{2y} \implies \frac{du}{dy} - u = e^{2y}$$ This is a linear DE in u$u$ with respect to y$y$. P(y) = -1, Q(y) = e^2y$P(y) = -1, Q(y) = e^{2y}$. Integrating Factor: I.F. = e^int -1 dy = e^-y$I.F. = e^{\int -1 dy} = e^{-y}$. Solution: u e^-y = int e^2y e^-y dy = int e^y dy = e^y + C$$u e^{-y} = \int e^{2y} e^{-y} dy = \int e^y dy = e^y + C$$ Substitute u = frac1tan x$u = \frac{1}{\tan x}$: frace^-ytan x = e^y + C$$\frac{e^{-y}}{\tan x} = e^y + C$$ Use given condition y(pi/4) = 0$y(\pi/4) = 0$: frace^0tan(pi/4) = e^0 + C implies 1 = 1 + C implies C = 0$$\frac{e^0}{\tan(\pi/4)} = e^0 + C \implies 1 = 1 + C \implies C = 0$$ Therefore, frace^-ytan x = e^y implies tan x = e^-2y$\frac{e^{-y}}{\tan x} = e^y \implies \tan x = e^{-2y}$. Evaluate at x = pi/6$x = \pi/6$, y = alpha$y = \alpha$: tan(pi/6) = e^-2alpha implies frac1sqrt3 = e^-2alpha$$\tan(\pi/6) = e^{-2\alpha} \implies \frac{1}{\sqrt{3}} = e^{-2\alpha}$$ e^2alpha = sqrt3 implies (e^2alpha)^4 = (sqrt3)^4 = 9$$e^{2\alpha} = \sqrt{3} \implies (e^{2\alpha})^4 = (\sqrt{3})^4 = 9$$ Thus, e^8alpha = 9$e^{8\alpha} = 9$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Core Logic Given DE: fracdxdy = fracxy left(lnleft(fracxyright) + 1right)$\frac{dx}{dy} = \frac{x}{y} \left(\ln\left(\frac{x}{y}\right) + 1\right)$ Let fracxy = t implies x = ty$\frac{x}{y} = t \implies x = ty$. Differentiating w.r.t y$y$: fracdxdy = t + yfracdtdy$$\frac{dx}{dy} = t + y\frac{dt}{dy}$$ ### Step 1: Substitution and Integration t + yfracdtdy = t(ln(t) + 1) = tln t + t$$t + y\frac{dt}{dy} = t(\ln(t) + 1) = t\ln t + t$$ yfracdtdy = tln t implies fracdttln t = fracdyy$$y\frac{dt}{dy} = t\ln t \implies \frac{dt}{t\ln t} = \frac{dy}{y}$$ Integrate both sides. Let ln t = p implies frac1t dt = dp$\ln t = p \implies \frac{1}{t} dt = dp$. int fracdpp = int fracdyy$$\int \frac{dp}{p} = \int \frac{dy}{y}$$ ln|p| = ln y + C implies ln|ln t| = ln y + C$$\ln|p| = \ln y + C \implies \ln|\ln t| = \ln y + C$$ lnleft|lnleft(fracxyright)right| = ln y + C$$\ln\left|\ln\left(\frac{x}{y}\right)\right| = \ln y + C$$ ### Step 2: Applying Boundary Conditions Given curve passes through (e, 1)$(e, 1)$: lnleft|lnleft(frace1right)right| = ln(1) + C implies C = 0$$\ln\left|\ln\left(\frac{e}{1}\right)\right| = \ln(1) + C \implies C = 0$$ lnleft|lnleft(fracxyright)right| = ln y$$\ln\left|\ln\left(\frac{x}{y}\right)\right| = \ln y$$ left|lnleft(fracxyright)right| = e^ln y = y$$\left|\ln\left(\frac{x}{y}\right)\right| = e^{\ln y} = y$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Core Logic fracdydx = fracfracsin xcos x + ysin x left(frac1cos x - fracsin^2 xcos xright) = fracsin x + ycos xsin x (1 - sin^2 x)$$\frac{dy}{dx} = \frac{\frac{\sin x}{\cos x} + y}{\sin x \left(\frac{1}{\cos x} - \frac{\sin^2 x}{\cos x}\right)} = \frac{\sin x + y\cos x}{\sin x (1 - \sin^2 x)}$$ fracdydx = fracsin x + ycos xsin x cos^2 x = sec^2 x + frac2ysin 2x$$\frac{dy}{dx} = \frac{\sin x + y\cos x}{\sin x \cos^2 x} = \sec^2 x + \frac{2y}{\sin 2x}$$ fracdydx - 2csc(2x)y = sec^2 x$$\frac{dy}{dx} - 2\csc(2x)y = \sec^2 x$$ ### Step 1: Integrating Factor This is an LDE of form fracdydx + Py = Q$\frac{dy}{dx} + Py = Q$. I.F. = e^int -2csc(2x) dx$$I.F. = e^{\int -2\csc(2x) dx}$$ Let 2x = t implies 2dx = dt$2x = t \implies 2dx = dt$. I.F. = e^-int csc t dt = e^-ln|tan(t/2)| = e^-ln|tan x| = frac1|tan x|$$I.F. = e^{-\int \csc t dt} = e^{-\ln|\tan(t/2)|} = e^{-\ln|\tan x|} = \frac{1}{|\tan x|}$$ ### Step 2: Solution of LDE y(I.F.) = int Q(I.F.) dx + C$$y(I.F.) = \int Q(I.F.) dx + C$$ yfrac1tan x = int sec^2 x frac1tan x dx + C$$y\frac{1}{\tan x} = \int \sec^2 x \frac{1}{\tan x} dx + C$$ Let tan x = t implies sec^2 x dx = dt$\tan x = t \implies \sec^2 x dx = dt$. yfrac1tan x = int fracdtt + C = ln|tan x| + C$$y\frac{1}{\tan x} = \int \frac{dt}{t} + C = \ln|\tan x| + C$$ y = tan x(ln|tan x| + C)$$y = \tan x(\ln|\tan x| + C)$$ ### Step 3: Boundary Value Given y(pi/4) = 2$y(\pi/4) = 2$: 2 = 1(ln 1 + C) implies C = 2$$2 = 1(\ln 1 + C) \implies C = 2$$ Thus, y = tan x (ln|tan x| + 2)$y = \tan x (\ln|\tan x| + 2)$. At x = pi/3$x = \pi/3$: y(pi/3) = sqrt3(lnsqrt3 + 2)$$y(\pi/3) = \sqrt{3}(\ln\sqrt{3} + 2)$$ ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations