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

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

Question

(A) The given differential equation is $x^{4}dy + (4x^{3}y + 2\sin x)dx = 0$. Rearranging the terms, we get $x^{4}dy + 4x^{3}ydx = -2\sin x dx$. This

Vedclass · Accepted Answer

$81$

Vedclass · Answer

(A) The given differential equation is $x^{4}dy + (4x^{3}y + 2\sin x)dx = 0$. Rearranging the terms, we get $x^{4}dy + 4x^{3}ydx = -2\sin x dx$. This can be written as $d(x^{4}y) = -2\sin x dx$. Integrating both sides, we get $\int d(x^{4}y) = \int -2\sin x dx$, which gives $x^{4}y = 2\cos x + C$. Given the condition $y(\frac{\pi}{2}) = 0$, we substitute $x = \frac{\pi}{2}$ and $y = 0$: $(\frac{\pi}{2})^{4}(0) = 2\cos(\frac{\pi}{2}) + C \Rightarrow 0 = 2(0) + C \Rightarrow C = 0$. So, the solution is $x^{4}y = 2\cos x$. Now, we need to find $\pi^{4}y(\frac{\pi}{3})$. At $x = \frac{\pi}{3}$, we have $(\frac{\pi}{3})^{4} y(\frac{\pi}{3}) = 2\cos(\frac{\pi}{3})$. Since $\cos(\frac{\pi}{3}) = \frac{1}{2}$, we get $\frac{\pi^{4}}{81} y(\frac{\pi}{3}) = 2(\frac{1}{2}) = 1$. Therefore, $\pi^{4} y(\frac{\pi}{3}) = 81$.

Let $y=y(x)$ be the solution of the differential equation $x^{4}dy + (4x^{3}y + 2\sin x)dx = 0$, $x>0$, $y(\frac{\pi}{2})=0$. Then $\pi^{4}y(\frac{\pi}{3})$ is equal to:

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} + (\sec x \operatorname{cosec} x) y = \cos^2 x$ is

Let $f:(0, \infty) \rightarrow \mathbb{R}$ be a differentiable function such that $f^{\prime}(x)=2-\frac{f(x)}{x}$ for all $x \in(0, \infty)$ and $f(1) \neq 1$. Then

The solution of the differential equation $(x + 2y^3)\frac{dy}{dx} - y = 0$ is

Solve the differential equation $\left[\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right] \frac{d x}{dy}=1$ where $x \neq 0$.

The integrating factor of $\frac{dy}{dx} + \frac{y}{x} = x^3 - 3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module