If frac{dy}{dx} + frac{y}{x} = x^2,then 2y(2) | vedclass

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

Question

(B) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y

Vedclass · Accepted Answer

$\frac{15}{4}$

Vedclass · Answer

(B) The given differential equation is $\frac{dy}{dx} + \frac{y}{x} = x^2$. This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$,where $P(x) = \frac{1}{x}$ and $Q(x) = x^2$. The integrating factor $(IF)$ is given by $IF = e^{\int P(x) dx} = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. The general solution is $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$. Substituting the values,we get $y \cdot x = \int x^2 \cdot x dx + C = \int x^3 dx + C = \frac{x^4}{4} + C$. Thus,$y = \frac{x^3}{4} + \frac{C}{x}$. Now,$y(2) = \frac{2^3}{4} + \frac{C}{2} = 2 + \frac{C}{2}$. And $y(1) = \frac{1^3}{4} + \frac{C}{1} = \frac{1}{4} + C$. Calculating $2y(2) - y(1) = 2(2 + \frac{C}{2}) - (\frac{1}{4} + C) = 4 + C - \frac{1}{4} - C = 4 - \frac{1}{4} = \frac{15}{4}$.

If $\frac{dy}{dx} + \frac{y}{x} = x^2$,then $2y(2) - y(1) =$

Similar Questions

If $y(x)$ satisfies the differential equation $y' + y = 2(\sin x + \cos x)$ and $y(0) = 1$,then

Find the solution of the differential equation given below:
$\frac{dy}{dx} + y \cdot \csc^2 (x) = \csc^2 (x) \cdot \cot (x)$

The particular solution of the differential equation $\sin^{2} y \frac{dx}{dy} + x = \cot y$ when $x = 0$ and $y = \frac{3\pi}{4}$ is

Let $y=y(x)$ be the solution of the differential equation $\cos x(3 \sin x+\cos x+3) dy = (1+y \sin x(3 \sin x+\cos x+3)) dx$; $0 \leq x \leq \frac{\pi}{2}, y(0)=0$. Then,$y\left(\frac{\pi}{3}\right)$ is equal to ..... .

Find the general solution of the differential equation: $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module