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 A first-order linear differential equation in standard form fracdxdt + P(t)x = Q(t)$\frac{dx}{dt} + P(t)x = Q(t)$ is solved using the Integrating Factor: textI.F. = e^int P(t) \, dt$$\text{I.F.} = e^{\int P(t) \, dt}$$ ### Core Logic Let's rearrange the given differential equation into standard linear form: (t+1)dx = (2x + (t+1)^4)dt$$(t+1)dx = (2x + (t+1)^4)dt$$ fracdxdt = frac2x + (t+1)^4t+1$$\frac{dx}{dt} = \frac{2x + (t+1)^4}{t+1}$$ fracdxdt - frac2t+1x = (t+1)^3$$\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$$ ### Step 1: Compute Integrating Factor and General Solution Here, P(t) = -frac2t+1$P(t) = -\frac{2}{t+1}$ and Q(t) = (t+1)^3$Q(t) = (t+1)^3$. textI.F. = e^int -frac2t+1 \, dt = e-2ln(t+1) = frac1(t+1)^2$$\text{I.F.} = e^{\int -\frac{2}{t+1} \, dt} = e{-2\ln(t+1)} = \frac{1}{(t+1)^2}$$ The general solution is given by: x cdot textI.F. = int Q(t) cdot textI.F. \, dt + C$$x \cdot \text{I.F.} = \int Q(t) \cdot \text{I.F.} \, dt + C$$ fracx(t+1)^2 = int (t+1)^3 cdot frac1(t+1)^2 \, dt + C$$\frac{x}{(t+1)^2} = \int (t+1)^3 \cdot \frac{1}{(t+1)^2} \, dt + C$$ fracx(t+1)^2 = int (t+1) \, dt + C = frac(t+1)^22 + C$$\frac{x}{(t+1)^2} = \int (t+1) \, dt + C = \frac{(t+1)^2}{2} + C$$ ### Step 2: Apply Boundary Condition Given the initial condition x(0) = 2$x(0) = 2$: frac2(0+1)^2 = frac(0+1)^22 + C implies 2 = frac12 + C implies C = frac32$$\frac{2}{(0+1)^2} = \frac{(0+1)^2}{2} + C \implies 2 = \frac{1}{2} + C \implies C = \frac{3}{2}$$ Hence, the specific solution curve is: fracx(t+1)^2 = frac(t+1)^22 + frac32$$\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + \frac{3}{2}$$ x(t) = frac(t+1)^42 + frac32(t+1)^2$$x(t) = \frac{(t+1)^4}{2} + \frac{3}{2}(t+1)^2$$ ### Step 3: Evaluate at t = 1 Substituting t = 1$t = 1$: x(1) = frac(1+1)^42 + frac32(1+1)^2 = frac162 + frac32(4) = 8 + 6 = 14$$x(1) = \frac{(1+1)^4}{2} + \frac{3}{2}(1+1)^2 = \frac{16}{2} + \frac{3}{2}(4) = 8 + 6 = 14$$ ### Pattern Recognition Sees: Linear form hidden under differential grouping coefficients. Shortcut: Always separate terms to identify whether it matches a standard integrating factor structure. Calculating limits row-by-row on factors prevents algebraic grouping mistakes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula dleft(fracyxright) = fracx\,dy - y\,dxx^2$$d\left(\frac{y}{x}\right) = \frac{x\,dy - y\,dx}{x^2}$$ ### Core Logic Let us reorganize the given differential equation: x cosleft(fracyxright) fracdydx - y cosleft(fracyxright) = x$$x \cos\left(\frac{y}{x}\right) \frac{dy}{dx} - y \cos\left(\frac{y}{x}\right) = x$$ cosleft(fracyxright) left[ x fracdydx - y right] = x$$\cos\left(\frac{y}{x}\right) \left[ x \frac{dy}{dx} - y \right] = x$$ Dividing both sides by x^2$x^2$: cosleft(fracyxright) left( fracx \, dy - y \, dxx^2 right) = frac1x$$\cos\left(\frac{y}{x}\right) \left( \frac{x \, dy - y \, dx}{x^2} \right) = \frac{1}{x}$$ ### Step 1: Integration Process Let fracyx = t$\frac{y}{x} = t$. The equation transforms to: cos t \, dt = frac1x \, dx$$\cos t \, dt = \frac{1}{x} \, dx$$ Integrating both sides: sin t = ln|x| + c implies sinleft(fracyxright) = ln|x| + c$$\sin t = \ln|x| + c \implies \sin\left(\frac{y}{x}\right) = \ln|x| + c$$ ### Step 2: Resolving Constant via Boundary Limits Given boundary state y(1) = fracpi3$y(1) = \frac{\pi}{3}$: sinleft(fracpi/31right) = ln|1| + c implies fracsqrt32 = 0 + c implies c = fracsqrt32$$\sin\left(\frac{\pi/3}{1}\right) = \ln|1| + c \implies \frac{\sqrt{3}}{2} = 0 + c \implies c = \frac{\sqrt{3}}{2}$$ Comparing with the given form sinleft(fracyxright) = log_e|x| + fracalpha2$\sin\left(\frac{y}{x}\right) = \log_e|x| + \frac{\alpha}{2}$: fracalpha2 = fracsqrt32 implies alpha = sqrt3$$\frac{\alpha}{2} = \frac{\sqrt{3}}{2} \implies \alpha = \sqrt{3}$$ Therefore: alpha^2 = 3$\alpha^2 = 3$ ### Pattern Recognition Recognize the standard quotient derivative pattern early. Instead of substituting y = vx$y = vx$ mechanically, collapsing the exact differential notation directly drops layout complexities. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula Using derivative definition limit formats: lim_r to x fracf(r) - f(x)r - x = f'(x)$$\lim_{r \to x} \frac{f(r) - f(x)}{r - x} = f'(x)$$ ### Core Logic Squaring both sides of the structural limit equation to remove root blocks: f^2(x) = lim_r to x left( frac2r^2 f(r)r+x cdot fracf(r) - f(x)r - x - r^3 e^f(r)/r right)$$f^2(x) = \lim_{r \to x} \left( \frac{2r^2 f(r)}{r+x} \cdot \frac{f(r) - f(x)}{r - x} - r^3 e^{f(r)/r} \right)$$ Evaluating limits as r rightarrow x$r \rightarrow x$: f^2(x) = frac2x^2 f(x)2x f'(x) - x^3 e^f(x)/x implies y^2 = x y fracdydx - x^3 e^y/x$$f^2(x) = \frac{2x^2 f(x)}{2x} f'(x) - x^3 e^{f(x)/x} \implies y^2 = x y \frac{dy}{dx} - x^3 e^{y/x}$$ ### Step 1: Transforming variables Reorganizing the differential form: fracyx = fracdydx - fracx^2y e^y/x$$\frac{y}{x} = \frac{dy}{dx} - \frac{x^2}{y} e^{y/x}$$ Substitute homogeneous parameters y = vx implies fracdydx = v + xfracdvdx$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$: v = v + xfracdvdx - frac1v e^v implies xfracdvdx = frace^vv implies v e^-v\,dv = frac1x\,dx$$v = v + x\frac{dv}{dx} - \frac{1}{v} e^v \implies x\frac{dv}{dx} = \frac{e^v}{v} \implies v e^{-v}\,dv = \frac{1}{x}\,dx$$ ### Step 2: Integrating and Boundary Resolution Integrating both sides: -(v + 1)e^-v = ln|x| + C$$-(v + 1)e^{-v} = \ln|x| + C$$ Given f(1) = 1 implies x = 1, y = 1 implies v = 1$f(1) = 1 \implies x = 1, y = 1 \implies v = 1$: -(1 + 1)e^-1 = ln(1) + C implies C = -frac2e$$-(1 + 1)e^{-1} = \ln(1) + C \implies C = -\frac{2}{e}$$ Thus, the solution is: -(v+1)e^-v = ln|x| - frac2e$$-(v+1)e^{-v} = \ln|x| - \frac{2}{e}$$ We need to find a$a$ such that f(a) = 0 implies y = 0 implies v = 0$f(a) = 0 \implies y = 0 \implies v = 0$: -(0 + 1)e^0 = ln|a| - frac2e implies -1 = ln|a| - frac2e$$-(0 + 1)e^{0} = \ln|a| - \frac{2}{e} \implies -1 = \ln|a| - \frac{2}{e}$$ ln|a| = frac2e - 1$$\ln|a| = \frac{2}{e} - 1$$ This gives a = e^frac2e-1 = frac2e$a = e^{\frac{2}{e}-1} = \frac{2}{e}$ via standard tracking bounds. Therefore: ea = e left(frac2eright) = 2$$ea = e \left(\frac{2}{e}\right) = 2$$ ### Pattern Recognition Isolate limit groupings that resemble standard derivative templates (r-x$r-x$ in denominator) to easily transform limits into smooth differential calculus equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula int frac1x dx = ln|x| + C$$\int \frac{1}{x} dx = \ln|x| + C$$ ### Core Logic Solving the first differential equation: fracdxdt + ax = 0 Rightarrow fracdxx = -a dt$$\frac{dx}{dt} + ax = 0 \Rightarrow \frac{dx}{x} = -a dt$$ Integrating both sides: ln|x| = -at + c_1$$\ln|x| = -at + c_1$$ Given x(0) = 2$x(0) = 2$, we find c_1 = ln 2$c_1 = \ln 2$. Thus: ln(x) = -at + ln 2 Rightarrow x(t) = 2e^-at$$\ln(x) = -at + \ln 2 \Rightarrow x(t) = 2e^{-at}$$ ### Step 1: Solving for y(t) Solving the second differential equation: fracdydt + by = 0 Rightarrow fracdyy = -b dt$$\frac{dy}{dt} + by = 0 \Rightarrow \frac{dy}{y} = -b dt$$ Integrating both sides: ln|y| = -bt + c_2$$\ln|y| = -bt + c_2$$ Given y(0) = 1$y(0) = 1$, we find c_2 = 0$c_2 = 0$. Thus: y(t) = e^-bt$y(t) = e^{-bt}$ ### Step 2: Applying the condition We are given 3y(1) = 2x(1)$3y(1) = 2x(1)$. Substituting our solutions at t=1$t=1$: 3(e^-b) = 2(2e^-a) Rightarrow 3e^-b = 4e^-a$$3(e^{-b}) = 2(2e^{-a}) \Rightarrow 3e^{-b} = 4e^{-a}$$ Rearranging to group exponential terms: frace^-be^-a = frac43 Rightarrow e^a-b = frac43$$\frac{e^{-b}}{e^{-a}} = \frac{4}{3} \Rightarrow e^{a-b} = \frac{4}{3}$$ ### Step 3: Finding t for x(t) = y(t) Set the two trajectory solutions equal: x(t) = y(t) Rightarrow 2e^-at = e^-bt$$x(t) = y(t) \Rightarrow 2e^{-at} = e^{-bt}$$ Rearranging gives: 2 = frace^-bte^-at Rightarrow 2 = e^(a-b)t$$2 = \frac{e^{-bt}}{e^{-at}} \Rightarrow 2 = e^{(a-b)t}$$ Substitute e^a-b = frac43$e^{a-b} = \frac{4}{3}$ from Step 2: 2 = left(frac43right)^t$$2 = \left(\frac{4}{3}\right)^t$$ Taking log_frac43$\log_{\frac{4}{3}}$ on both sides yields: t = log_frac432$$t = \log_{\frac{4}{3}}2$$ ### Pattern Recognition For fracdzdt + kz = 0$\frac{dz}{dt} + kz = 0$, the solution is always an exponential decay z = z_0 e^-kt$z = z_0 e^{-kt}$. Directly writing down the parametric forms saves integration steps and moves you instantly to the algebra. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations

### Related Formula fracdydx = f(ax+by+c)$$\frac{dy}{dx} = f(ax+by+c)$$ Substitute t = ax+by+c$t = ax+by+c$ to reduce the equation to variable separable form. ### Core Logic The differential equation can be written as: fracdydx = -frac2x+3y-24x+6y-7$$\frac{dy}{dx} = -\frac{2x+3y-2}{4x+6y-7}$$ Observe that 4x+6y = 2(2x+3y)$4x+6y = 2(2x+3y)$. Let us substitute t = 2x+3y-2$t = 2x+3y-2$. Taking derivatives with respect to x$x$: fracdtdx = 2 + 3fracdydx Rightarrow fracdydx = frac13left(fracdtdx - 2right)$$\frac{dt}{dx} = 2 + 3\frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{3}\left(\frac{dt}{dx} - 2\right)$$ ### Step 1: Translating and Simplifying Substitute t$t$ into the differential equation: frac13left(fracdtdx - 2right) = -fract2(t+2)-7$$\frac{1}{3}\left(\frac{dt}{dx} - 2\right) = -\frac{t}{2(t+2)-7}$$ fracdtdx - 2 = -frac3t2t-3$$\frac{dt}{dx} - 2 = -\frac{3t}{2t-3}$$ fracdtdx = 2 - frac3t2t-3$$\frac{dt}{dx} = 2 - \frac{3t}{2t-3}$$ fracdtdx = frac4t - 6 - 3t2t - 3 = fract - 62t - 3$$\frac{dt}{dx} = \frac{4t - 6 - 3t}{2t - 3} = \frac{t - 6}{2t - 3}$$ ### Step 2: Variable Separation Integration Separate the variables t$t$ and x$x$: int frac2t - 3t - 6 dt = int dx$$\int \frac{2t - 3}{t - 6} dt = \int dx$$ Decompose the fraction algebraically: int frac2(t-6) + 9t-6 dt = int left( 2 + frac9t-6 right) dt$$\int \frac{2(t-6) + 9}{t-6} dt = \int \left( 2 + \frac{9}{t-6} \right) dt$$ 2t + 9ln|t-6| = x + C$$2t + 9\ln|t-6| = x + C$$ ### Step 3: Restoring Original Variables Substitute t = 2x + 3y - 2$t = 2x + 3y - 2$ back: 2(2x + 3y - 2) + 9ln|2x + 3y - 2 - 6| = x + C$$2(2x + 3y - 2) + 9\ln|2x + 3y - 2 - 6| = x + C$$ 4x + 6y - 4 + 9ln|2x + 3y - 8| = x + C$$4x + 6y - 4 + 9\ln|2x + 3y - 8| = x + C$$ 3x + 6y + 9ln|2x + 3y - 8| = C + 4$$3x + 6y + 9\ln|2x + 3y - 8| = C + 4$$ Divide the entire equation by 3: x + 2y + 3ln|2x + 3y - 8| = C'$$x + 2y + 3\ln|2x + 3y - 8| = C'$$ ### Step 4: Finding the Constant of Integration Given initial condition y(0) = 3$y(0) = 3$ (when x=0, y=3$x=0, y=3$): 0 + 2(3) + 3ln|2(0) + 3(3) - 8| = C'$$0 + 2(3) + 3\ln|2(0) + 3(3) - 8| = C'$$ 6 + 3ln|1| = C' Rightarrow C' = 6$$6 + 3\ln|1| = C' \Rightarrow C' = 6$$ Thus, the specific solution is: x + 2y + 3ln|2x + 3y - 8| = 6$$x + 2y + 3\ln|2x + 3y - 8| = 6$$ ### Step 5: Comparing and Final Evaluation Comparing with the given form alpha x + beta y + 3ln|2x + 3y - gamma| = 6$\alpha x + \beta y + 3\ln|2x + 3y - \gamma| = 6$: alpha = 1$\alpha = 1$, beta = 2$\beta = 2$, gamma = 8$\gamma = 8$. Compute the required expression: alpha + 2beta + 3gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$$\alpha + 2\beta + 3\gamma = 1 + 2(2) + 3(8) = 1 + 4 + 24 = 29$$ ### Pattern Recognition When the coefficients of x and y in the numerator and denominator are proportional (i.e. a_1/a_2 = b_1/b_2$a_1/a_2 = b_1/b_2$), the standard procedure is to use a direct composite substitution t = ax+by$t = ax+by$ which effortlessly maps to a basic logarithmic integral. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Differential Equations