### 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: textI.F. = e^int P(x) dx$\text{I.F.} = e^{\int P(x) dx}$ ### Core Logic Rearrange the given differential equation: sqrt4-x^2 fracdydx = left( sin^-1left(fracx2right) right)^3 - y sin^-1left(fracx2 ight)$\sqrt{4-x^2} \frac{dy}{dx} = \left( \sin^{-1}\left(\frac{x}{2}\right) \right)^3 - y \sin^{-1}\left(\frac{x}{2} ight)$ Divide both sides by sqrt4-x^2$\sqrt{4-x^2}$: fracdydx + fracsin^-1(x/2)sqrt4-x^2 y = frac(sin^-1(x/2))^3sqrt4-x^2$\frac{dy}{dx} + \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} y = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$ ### Step 1: Calculate Integrating Factor and Solve textI.F. = e^int fracsin^-1(x/2)sqrt4-x^2 dx = e^frac12 (sin^-1(x/2))^2$\text{I.F.} = e^{\int \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} dx} = e^{\frac{1}{2} (\sin^{-1}(x/2))^2}$ The general solution follows the structure: y = left( sin^-1left(fracx2right) right)^2 - 2 + C cdot e^-frac12 (sin^-1(x/2))^2$y = \left( \sin^{-1}\left(\frac{x}{2}\right) \right)^2 - 2 + C \cdot e^{-\frac{1}{2} (\sin^{-1}(x/2))^2}$ Using the initial condition y(2) = fracpi^2 - 84 = fracpi^24 - 2$y(2) = \frac{\pi^2 - 8}{4} = \frac{\pi^2}{4} - 2$: fracpi^24 - 2 = left(fracpi2right)^2 - 2 + C cdot e^-fracpi^28 implies C = 0$\frac{\pi^2}{4} - 2 = \left(\frac{\pi}{2}\right)^2 - 2 + C \cdot e^{-\frac{\pi^2}{8}} \implies C = 0$ Thus, the specific solution simplifies to: y(x) = left( sin^-1left(fracx2right) right)^2 - 2$y(x) = \left( \sin^{-1}\left(\frac{x}{2}\right) \right)^2 - 2$ ### Step 2: Evaluate at x=0 $y(0) = (sin^-1(0))^2 - 2 = 0 - 2 = -2$$y(0) = (\sin^{-1}(0))^2 - 2 = 0 - 2 = -2$ Therefore: y^2(0) = (-2)^2 = 4$y^2(0) = (-2)^2 = 4$ ### Pattern Recognition Recognizing that the coefficient of y$y$ is precisely the derivative of frac12(sin^-1(x/2))^2$\frac{1}{2}(\sin^{-1}(x/2))^2$ makes computing the linear integrating factor simple. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula textStandard Linear Form: fracdydx + P(x)y = Q(x)$\text{Standard Linear Form: } \frac{dy}{dx} + P(x)y = Q(x)$ textIntegrating Factor (I.F.) = e^int P(x) \, dx$\text{Integrating Factor (I.F.)} = e^{\int P(x) \, dx}$ ### Core Logic Rearranging the equation to standard linear differential form: cos x (ln(cos x))^2 fracdydx - 3sin x ln(cos x)y = -sin x$\cos x (\ln(\cos x))^2 \frac{dy}{dx} - 3\sin x \ln(\cos x)y = -\sin x$ Divide by cos x (ln(cos x))^2$\cos x (\ln(\cos x))^2$: fracdydx - frac3tan xln(cos x)y = frac-tan x(ln(cos x))^2$\frac{dy}{dx} - \frac{3\tan x}{\ln(\cos x)}y = \frac{-\tan x}{(\ln(\cos x))^2}$ Alternatively, writing in terms of sec x$\sec x$: fracdydx + frac3tan xln(sec x)y = frac-tan x(ln(sec x))^2$\frac{dy}{dx} + \frac{3\tan x}{\ln(\sec x)}y = \frac{-\tan x}{(\ln(\sec x))^2}$ ### Step 1: Compute Integrating Factor I.F. = e^int frac3tan xln(sec x) \, dx = e^3ln(ln(sec x)) = (ln(sec x))^3$I.F. = e^{\int \frac{3\tan x}{\ln(\sec x)} \, dx} = e^{3\ln(\ln(\sec x))} = (\ln(\sec x))^3$ ### Step 2: Solve the Integral Solution y times (ln(sec x))^3 = -int fractan x(ln(sec x))^2 (ln(sec x))^3 \, dx$y \times (\ln(\sec x))^3 = -\int \frac{\tan x}{(\ln(\sec x))^2} (\ln(\sec x))^3 \, dx$ y times (ln(sec x))^3 = -int tan x ln(sec x) \, dx = -frac12(ln(sec x))^2 + C$y \times (\ln(\sec x))^3 = -\int \tan x \ln(\sec x) \, dx = -\frac{1}{2}(\ln(\sec x))^2 + C$ ### Step 3: Apply Boundary Condition Given x = fracpi4$x = \frac{\pi}{4}$, y = -frac1ln 2$y = -\frac{1}{\ln 2}$. Note sec(fracpi4) = sqrt2$\sec(\frac{\pi}{4}) = \sqrt{2}$. left(-frac1ln 2right) left(lnsqrt2right)^3 = -frac12left(lnsqrt2right)^2 + C$\left(-\frac{1}{\ln 2}\right) \left(\ln\sqrt{2}\right)^3 = -\frac{1}{2}\left(\ln\sqrt{2}\right)^2 + C$ left(-frac1ln 2right) left(frac12ln 2right)^3 = -frac12left(frac12ln 2right)^2 + C implies C = 0$\left(-\frac{1}{\ln 2}\right) \left(\frac{1}{2}\ln 2\right)^3 = -\frac{1}{2}\left(\frac{1}{2}\ln 2\right)^2 + C \implies C = 0$ ### Step 4: Final Substitution for x = pi/6$x = \pi/6$ With C=0$C=0$, y = frac-12ln(sec x) = frac12ln(cos x)$y = \frac{-1}{2\ln(\sec x)} = \frac{1}{2\ln(\cos x)}$. yleft(fracpi6right) = frac12lnleft(fracsqrt32right) = frac12left(frac12ln 3 - ln 2right) = frac1ln 3 - ln 4$y\left(\frac{\pi}{6}\right) = \frac{1}{2\ln\left(\frac{\sqrt{3}}{2}\right)} = \frac{1}{2\left(\frac{1}{2}\ln 3 - \ln 2\right)} = \frac{1}{\ln 3 - \ln 4}$ ### Pattern Recognition Logarithmic functions nested inside trigonometric terms generally point to substitution structures where ln(sec x)$\ln(\sec x)$ works cleanly alongside its derivative tan x \, dx$\tan x \, dx$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### Related Formula For differential equations where terms are functions of a linear expression like (x+y)$(x+y)$, substitute a new variable t = x+y$t = x+y$ to enable variable separation. ### Core Logic Given the differential equation: fracdydx = 2x(x+y)^3 - x(x+y) - 1$\frac{dy}{dx} = 2x(x+y)^3 - x(x+y) - 1$ Let x+y = t implies 1 + fracdydx = fracdtdx implies fracdydx = fracdtdx - 1$x+y = t \implies 1 + \frac{dy}{dx} = \frac{dt}{dx} \implies \frac{dy}{dx} = \frac{dt}{dx} - 1$. Substituting these terms back: fracdtdx - 1 = 2xt^3 - xt - 1$\frac{dt}{dx} - 1 = 2xt^3 - xt - 1$ fracdtdx = 2xt^3 - xt implies fracdtdx = xt(2t^2 - 1)$\frac{dt}{dx} = 2xt^3 - xt \implies \frac{dt}{dx} = xt(2t^2 - 1)$ ### Step 1: Separating Variables and Integrating Separating variables: fracdtt(2t^2 - 1) = x \, dx$\frac{dt}{t(2t^2 - 1)} = x \, dx$ Multiply numerator and denominator by t$t$: fract \, dtt^2(2t^2 - 1) = x \, dx$\frac{t \, dt}{t^2(2t^2 - 1)} = x \, dx$ Let t^2 = z implies 2t \, dt = dz implies t \, dt = fracdz2$t^2 = z \implies 2t \, dt = dz \implies t \, dt = \frac{dz}{2}$: int fracdz2z(2z-1) = int x \, dx implies int fracdzz(2z-1) = int 2x \, dx$\int \frac{dz}{2z(2z-1)} = \int x \, dx \implies \int \frac{dz}{z(2z-1)} = \int 2x \, dx$ Using partial fractions: int left( frac22z-1 - frac1z right) dz = int 2x \, dx$\int \left( \frac{2}{2z-1} - \frac{1}{z} \right) dz = \int 2x \, dx$ ln|2z-1| - ln|z| = x^2 + C implies lnleft|frac2z-1zright| = x^2 + C$\ln|2z-1| - \ln|z| = x^2 + C \implies \ln\left|\frac{2z-1}{z}\right| = x^2 + C$ ### Step 2: Apply the Boundary Condition Given y(0) = 1 implies$y(0) = 1 \implies$ at x = 0$x = 0$, y = 1 implies t = 0 + 1 = 1 implies z = t^2 = 1$y = 1 \implies t = 0 + 1 = 1 \implies z = t^2 = 1$. Substituting these value constraints into our integral solution: lnleft|frac2(1)-11right| = 0^2 + C implies ln(1) = C implies C = 0$\ln\left|\frac{2(1)-1}{1}\right| = 0^2 + C \implies \ln(1) = C \implies C = 0$ Thus: frac2z-1z = e^x^2 implies 2 - frac1z = e^x^2 implies frac1z = 2 - e^x^2 implies z = frac12 - e^x^2$\frac{2z-1}{z} = e^{x^2} \implies 2 - \frac{1}{z} = e^{x^2} \implies \frac{1}{z} = 2 - e^{x^2} \implies z = \frac{1}{2 - e^{x^2}}$ Since z = t^2 = (x+y)^2$z = t^2 = (x+y)^2$, we have: (x+y)^2 = frac12 - e^x^2$(x+y)^2 = \frac{1}{2 - e^{x^2}}$ ### Step 3: Evaluate at the Target Value We need to find the value of the function at x = frac1sqrt2$x = \frac{1}{\sqrt{2}}$: left( frac1sqrt2 + yleft(frac1sqrt2right) right)^2 = frac12 - e^left(frac1sqrt2right)^2 = frac12 - e^1/2 = frac12 - sqrte$\left( \frac{1}{\sqrt{2}} + y\left(\frac{1}{\sqrt{2}}\right) \right)^{2} = \frac{1}{2 - e^{\left(\frac{1}{\sqrt{2}}\right)^2}} = \frac{1}{2 - e^{1/2}} = \frac{1}{2 - \sqrt{e}}$ ### Pattern Recognition Sees: Differential equation format y' = f(x+y)$y' = f(x+y)$. Shortcut: A linear argument (x+y)$(x+y)$ strongly implies substituting t=x+y$t=x+y$. The final expression requested matched the functional template (x+y)^2$(x+y)^2$ exactly, saving steps from extracting standalone square roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Differential Equations

### 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