Let y = y(x)$y = y(x)$ be the solution curve of the differential equation (1 + x²)dy + (y - ⁻¹x)dx = 0$(1 + x^2)dy + (y - \tan^{-1}x)dx = 0$ , y(0) = 1$y(0) = 1$ . Then the value of y(1)$y(1)$ is:

Related Formula

For a quadratic equation Am² + Bm + C = 0$Am^2 + Bm + C = 0$ having equal roots, Discriminant D = 0 ⇒ B² - 4AC = 0$D = 0 \Rightarrow B^2 - 4AC = 0$.

Core Logic

Given quadratic equation in m$m$: f(x)m² - 2f'(x)m + f''(x) = 0$f(x)m^2 - 2f'(x)m + f''(x) = 0$ has equal roots.

Step 1: Solve the Differential Equation

Rewrite the DE: (f''(x))/(f'(x)) = (f'(x))/(f(x))$\frac{f''(x)}{f'(x)} = \frac{f'(x)}{f(x)}$ Integrate both sides:

Using given f(0) = 1$f(0) = 1$ and f'(0) = 2$f'(0) = 2$:

Now we have f'(x) = 2f(x) ⇒ (f'(x))/(f(x)) = 2$f'(x) = 2f(x) \Rightarrow \frac{f'(x)}{f(x)} = 2$. Integrate again:

Use f(0) = 1 ⇒ ln(1) = 0 + d ⇒ d = 0$f(0) = 1 \Rightarrow \ln(1) = 0 + d \Rightarrow d = 0$. So, ln f(x) = 2x ⇒ f(x) = e2x$\ln f(x) = 2x \Rightarrow f(x) = e^{2x}$.

Step 2: Investigate increasing interval

Let g(x) = f(ln x - x) = e2(ln x - x)$g(x) = f(\ln x - x) = e^{2(\ln x - x)}$. For g(x)$g(x)$ to be increasing, g'(x) ≥ 0$g'(x) \geq 0$.

Since exponential is always positive, g'(x) ≥ 0 ⇒ (1)/(x) - 1 ≥ 0$g'(x) \geq 0 \Rightarrow \frac{1}{x} - 1 \geq 0$.

The critical points are x=0, x=1$x=0, x=1$. Based on domain of ln x$\ln x$, x > 0$x > 0$. Sign scheme yields positive derivative in x in (0, 1]$x \in (0, 1]$. Therefore, (α, β) = (0, 1) ⇒ α = 0, β = 1$(\alpha, \beta) = (0, 1) \Rightarrow \alpha = 0, \beta = 1$.

Step 3: Final Output

Pattern Recognition

The relation (f')² = f · f''$(f')^2 = f \cdot f''$ is a classical indicator of exponential functions (f = Cekx$f = Ce^{kx}$). Solving via double logarithmic integration collapses the differential equation almost instantaneously.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Applications of Derivatives

Related Formula

Core Logic

Multiply the entire differential equation by x$\cos x$ to normalize the (dy)/(dx)$\frac{dy}{dx}$ term:

This is a standard first-order linear differential equation with P(x) = -2 x$P(x) = -2\cos x$.

Step 1: Integrating Factor

Step 2: General Solution

Let u = x du = x dx$u = \sin x \implies du = \cos x \, dx$. The integral becomes:

Using integration by parts:

Re-substitute u = x$u = \sin x$:

Step 3: Apply Boundary Conditions

Given y(0) = -(7)/(4)$y(0) = -\frac{7}{4}$:

Thus, y(x) = -(3)/(2) x - (7)/(4)$y(x) = -\frac{3}{2}\sin x - \frac{7}{4}$.

Step 4: Find Final Value

Calculate y((π)/(6))$y\left(\frac{\pi}{6}\right)$:

Pattern Recognition

When integrating eax(f(x))$e^{ax}(f(x))$, use DI method or standard Integration by Parts mapping. For ∫ e-2u(3u+2)$\int e^{-2u}(3u+2)$, always extract polynomial as u$u$, exponential as dv$dv$.

Chapter Mix

Class 12 Maths: Differential Equations

Related Formula

Core Logic

Given 6∫₁xf(t)dt=3xf(x)+x³-4$6\int_{1}^{x}f(t)dt=3xf(x)+x^{3}-4$. Differentiating both sides with respect to x$x$:

Step 1: Solving the Differential Equation

Let y = f(x)$y = f(x)$. The equation simplifies to:

Divide the entire equation by x²$x^2$ to recognize the exact differential form:

Integrate both sides:

Step 2: Evaluating the Constant

To find C$C$, we plug x=1$x=1$ into the original integral equation:

Using f(x) = -x² + Cx$f(x) = -x^2 + Cx$:

Thus, f(x) = -x² + 2x$f(x) = -x^2 + 2x$.

Step 3: Final Calculation

We need to find f(2) - f(3)$f(2) - f(3)$:

Pattern Recognition

An integral equation involving a generic function f(t)$f(t)$ within variable limits signals the immediate use of Leibniz's rule. Differentiating leads cleanly to a linear differential equation. Re-substituting the lower limit (x=1$x=1$) securely yields the boundary condition f(1)$f(1)$.

Chapter Mix

Class 12 Maths: Differential Equations Class 12 Maths: Applications of Integrals

Related Formula

Core Logic

Divide the entire differential equation by x²$x^2$ to construct an exact differential on the left hand side.

Rewrite the right side by pushing an x$x$ under the square root:

Step 1: Integration

Isolate the variables by dividing by the square root term:

Applying standard integral ∫ du√(1+u²) = ln(u + √(1+u²))$\int \frac{du}{\sqrt{1+u^2}} = \ln(u + \sqrt{1+u^2})$:

Step 2: Finding Constant of Integration

Remove logarithms:

Apply the initial condition y(1) = 0$y(1) = 0$:

The specific curve is y + √(x² + y²) = x²$y + \sqrt{x^2 + y^2} = x^2$.

Step 3: Calculating Final Value

Substitute x = 3$x = 3$ to find y(3)$y(3)$:

Square both sides:

Pattern Recognition

The expression x dy - y dx$x \, dy - y \, dx$ is the universal flag for the quotient rule exact differential d(y/x)$d(y/x)$. Dividing the whole equation by x²$x^2$ immediately separates variables.

Chapter Mix

Class 12 Maths: Differential Equations