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

Let $f(x)=x-1$ and $g(x)=e^x$ for $x \in R$. If $\frac{d y}{d x}=\left(e^{-2 \sqrt{x}} g(f(f(x)))-\frac{y}{\sqrt{x}}\right)$ and $y(0)=0$,then $y(1)$ is:

The solution of the differential equation $\frac{dy}{dx} + y \sec^2 x = \tan x \sec^2 x$ is

Let $y = y(x)$ be the solution of the differential equation,$x\frac{dy}{dx} + y = x \ln x$,for $x > 1$. If $2y(2) = \ln 4 - 1$,then $y(e)$ is equal to

The integrating factor of the differential equation $(2x + 3y^2) dy = y dx$ $(y > 0)$ is

If $\int_{a}^{x} t y(t) dt = x^2 + y(x)$,then $y$ as a function of $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module