If for the solution curve y = f(x)$y = f(x)$ of the differential equation (dy)/(dx) + ( x)y = (2 + x)/((1 + 2 x)²)$\frac{dy}{dx} + (\tan x)y = \frac{2 + \sec x}{(1 + 2\sec x)^2}$, x in ((-π)/(2), (π)/(2)), f((π)/(3)) = √(3)10$x \in \left(\frac{-\pi}{2}, \frac{\pi}{2}\right), f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{10}$, then f((π)/(4))$f\left(\frac{\pi}{4}\right)$ is equal to:

Related Formula

Leibniz Rule for differentiating under the integral sign:

Core Logic

Differentiate both sides of the integral equation with respect to x$x$:

Step 1: Solving the Separable Differential Equation

Separating variables:

Step 2: Applying Boundary Constraints

Given f(2) = 3$f(2) = 3$:

Evaluating for x = 6$x = 6$:

Pattern Recognition

Differentiating integral statements instantly converts complex integral equations into clean, separable differential equations.

Chapter Mix

Class 12 Maths: Differential Equations

Related Formula

Leibniz Integral Rule for differentiation:

Core Logic

Differentiate the given integral relation using Leibniz rule [cite: 1344]: ddx[∫₀xg(t)dt] = ddx[x-∫₀xtg(t)dt]$$\frac{\mathrm{d}}{\mathrm{d}x}\left[\int_{0}^{x}g(t)dt\right] = \frac{\mathrm{d}}{\mathrm{d}x}\left[x-\int_{0}^{x}tg(t)dt\right]$$ [cite: 1344] g(x) = 1 - xg(x) g(x)(1+x) = 1 g(x) = (1)/(1+x)$$g(x) = 1 - xg(x) \implies g(x)(1+x) = 1 \implies g(x) = \frac{1}{1+x}$$ [cite: 1345]

Substitute g(x)$g(x)$ into target differential equation configuration [cite: 1346]: dydx - y x = 2(x+1) x · ((1)/(1+x)) = 2 x$$\frac{\mathrm{d}y}{\mathrm{d}x} - y\tan x = 2(x+1)\sec x \cdot \left(\frac{1}{1+x}\right) = 2\sec x$$ [cite: 1346]

Step 1: Finding the Integrating Factor

This matches a linear form dydx + P(x)y = Q(x)$\frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y = Q(x)$ where P(x) = - x$P(x) = -\tan x$. I.F. = e∫ - x dx = eln| x| = x$$\text{I.F.} = e^{\int -\tan x \, \mathrm{d}x} = e^{\ln|\cos x|} = \cos x$$ [cite: 1346]

Write general functional solution template [cite: 1348]: y · x = ∫ (2 x · x) dx = ∫ 2 dx = 2x + C$$y \cdot \cos x = \int (2\sec x \cdot \cos x) \, \mathrm{d}x = \int 2 \, \mathrm{d}x = 2x + C$$ [cite: 1348]

Given boundary condition y(0) = 0 0 = 0 + C C = 0$y(0) = 0 \implies 0 = 0 + C \implies C = 0$ [cite: 1349]. y(x) = (2x)/( x) = 2x x$$y(x) = \frac{2x}{\cos x} = 2x\sec x$$ [cite: 1350]

Step 2: Numeric substitution

Substitute variable parameter values x = (π)/(3)$x = \frac{\pi}{3}$ [cite: 1352]: y((π)/(3)) = 2((π)/(3)) ((π)/(3)) = (2π)/(3) · 2 = (4π)/(3)$$y\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) = \frac{2\pi}{3} \cdot 2 = \frac{4\pi}{3}$$ [cite: 1351]

Pattern Recognition

Integral functional definitions are codes for simpler underlying derivatives. Applying Leibniz rule immediately extracts the true variable functions.

Chapter Mix

Class 12 Mathematics: Differential Equations

Core Logic

Let's rearrange the given differential equation by expressing (dy)/(dx)$\frac{dy}{dx}$:

Dividing both sides by x⁵$x^5$:

Multiply the entire equation by y$\sin y$ to clear denominators:

This can be transformed into a linear form by substituting t = - y$t = -\cos y$. Then (dt)/(dx) = y (dy)/(dx)$\frac{dt}{dx} = \sin y \frac{dy}{dx}$.

Step 1: Solving the Linear ODE

Substituting t$t$ leads to:

This is a standard linear first-order ODE with P(x) = (7)/(x)$P(x) = \frac{7}{x}$. The Integrating Factor (I.F.) is:

The general solution is:

Step 2: Evaluating the Integration and Constant

Using integration by parts for ∫ x² e^x dx$\int x^2 e^x dx$:

Substituting this back:

Since the curve passes through (1, (π)/(2))$\left(1, \frac{\pi}{2}\right)$:

Step 3: Calculating cos y at x = 2

Now substitute x = 2$x = 2$ and C = e$C = e$ into our equation block:

Pattern Recognition

When trigonometric terms are mixed inside an ODE containing derivative blocks like (dx)/(dy)$\frac{dx}{dy}$ or (dy)/(dx)$\frac{dy}{dx}$, check if clearing denominators using y$\sin y$ or y$\cos y$ reveals a standard substitution path for a Linear ODE.

Chapter Mix

Class 12 Mathematics: Differential Equations

Related Formula

Leibniz Integral Rule for differentiation under integral sign:

Core Logic

Rewrite equation to isolate the integral kernel:

Differentiating with respect to x$x$ using product rule and Leibniz rule:

Notice that e^x ∫₀^x e-t f(t)dt = y - (1 - 2x)$e^x \int_0^x e^{-t} f(t)\mathrm{d}t = y - (1 - 2x)$. Substitute this back:

Step 1: Solve Differential Equation

Integrating factor I.F. = e∫ -2 dx = e-2x$\text{I.F.} = e^{\int -2 \mathrm{d}x} = e^{-2x}$.

From original equation, at x=0$x=0$, f(0) = 1$f(0) = 1$. Evaluating c$c$:

Thus, y = -x + 1 x + y = 1$y = -x + 1 \implies x + y = 1$.

Step 2: Calculate Area

The boundary line is x + y = 1$x + y = 1$. The area bounded by this line and coordinate axes is a right triangle with intercepts (1,0)$(1,0)$ and (0,1)$(0,1)$:

Pattern Recognition

Integral equations of convolution type (ex-t$e^{x-t}$) always simplify directly into standard linear ordinary differential equations of first or second order when tracking Leibniz rules properly.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Area Under Curves

Related Formula

For a linear differential equation (dy)/(dx) + Py = Q$\frac{dy}{dx} + Py = Q$, the Integrating Factor (IF) is defined as:

Core Logic

Divide the full differential equation by (x²+1)$(x^2+1)$:

This is a standard Linear Differential Equation with:

Step 1: Solve for General Solution

The solution format is y · IF = ∫ Q · IF dx$y \cdot \text{IF} = \int Q \cdot \text{IF} \, dx$:

Using the boundary condition y(0) = 1$y(0) = 1$:

Step 2: Definite Integration Evaluation

We need to evaluate ∫₋₃³ y dx$\int_{-3}^{3} y \, dx$:

By symmetry of odd/even functions over symmetric intervals [-a, a]$[-a, a]$: ∫₋₃³ x² x dx = 0$\int_{-3}^{3} x^2\sin x \, dx = 0$ (since it is an odd function) ∫₋₃³ x dx = 0$\int_{-3}^{3} \sin x \, dx = 0$ (since it is an odd function)

Thus, we are left with the even components:

Pattern Recognition

Splitting a symmetric interval integral into odd and even parts immediately simplifies calculations by dropping all odd functions down to zero.

Chapter Mix

Class 12 Mathematics: Differential Equations Class 12 Mathematics: Integral Calculus