Let y=y(x) be the solution curve of the diffe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the differential equation: $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$. Divide both sides by $\cos y$ (or multiply by $\sec y$): $\sec^2

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Given the differential equation: $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$. Divide both sides by $\cos y$ (or multiply by $\sec y$): $\sec^2 y \frac{dy}{dx} + 2x 	an y = x^3$. Let $t = 	an y$,then $\frac{dt}{dx} = \sec^2 y \frac{dy}{dx}$. The equation becomes a linear differential equation: $\frac{dt}{dx} + 2xt = x^3$. The integrating factor $IF = e^{\int 2x dx} = e^{x^2}$. Multiplying by $IF$: $\frac{d}{dx}(t e^{x^2}) = x^3 e^{x^2}$. Integrating both sides: $t e^{x^2} = \int x^3 e^{x^2} dx + C$. Let $u = x^2$,then $du = 2x dx$,so $\int x^3 e^{x^2} dx = \frac{1}{2} \int u e^u du = \frac{1}{2} (u e^u - e^u) + C = \frac{1}{2} e^{x^2} (x^2 - 1) + C$. Thus,$	an y \cdot e^{x^2} = \frac{1}{2} e^{x^2} (x^2 - 1) + C$. Given $y(1) = 0$,so $	an(0) \cdot e^1 = \frac{1}{2} e^1 (1 - 1) + C \implies 0 = 0 + C \implies C = 0$. So,$	an y = \frac{1}{2} (x^2 - 1)$. For $x = \sqrt{3}$,$	an y = \frac{1}{2} ((\sqrt{3})^2 - 1) = \frac{1}{2} (3 - 1) = 1$. Since $	an y = 1$,$y = \frac{\pi}{4}$.

Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$,with the initial condition $y(1) = 0$. Then the value of $y(\sqrt{3})$ is equal to:

Similar Questions

The Integrating Factor of the differential equation $x \frac{dy}{dx} - y = 2x^2$ is

General solution of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is

The general solution of the differential equation $x^{2} dy - 2xy dx = x^{4} \cos x dx$ is

Integrating factor of $x \frac{dy}{dx} - y = x^4 - 3x$ is

The equation of the curve passing through $(1,2)$ and whose tangent at any point $(x, y)$ makes an angle $\tan ^{-1}(2 x+3 y)$ with the $X$-axis is .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module