If frac{dy}{dx} = y + 3,y + 3 0 and y(0) = | vedclass

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

Question

(C) Given the differential equation $\frac{dy}{dx} = y + 3$. Separating the variables,we get $\frac{dy}{y + 3} = dx$. Integrating both sides,we have $

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Given the differential equation $\frac{dy}{dx} = y + 3$. Separating the variables,we get $\frac{dy}{y + 3} = dx$. Integrating both sides,we have $\int \frac{dy}{y + 3} = \int dx + C$. This gives $\log(y + 3) = x + C$ ... $(i)$. Given $y(0) = 2$,substitute $x = 0$ and $y = 2$ into $(i)$: $\log(2 + 3) = 0 + C \Rightarrow C = \log 5$. Substituting $C$ back into $(i)$,we get $\log(y + 3) = x + \log 5$. Taking the exponential of both sides,$y + 3 = e^{x + \log 5} = 5e^x$. Thus,$y = 5e^x - 3$. Now,calculate $y(\log 2)$: $y(\log 2) = 5e^{\log 2} - 3 = 5(2) - 3 = 10 - 3 = 7$.

If $\frac{dy}{dx} = y + 3$,$y + 3 > 0$ and $y(0) = 2$,then $y(\log 2)$ is equal to

Similar Questions

The general solution of $\frac{dy}{dx} + \sin \left(\frac{x+y}{2}\right) = \sin \left(\frac{x-y}{2}\right)$ is

The general solution of the differential equation ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$ is

The general solution of the differential equation $\tan x \tan y \, dx + \cos^2 x \operatorname{cosec}^2 y \, dy = 0$ is

The solution of the differential equation $(2x - 3y + 5)dx + (9y - 6x - 7)dy = 0$ is

If $x \frac{dy}{dx} + y = x \frac{f(xy)}{f'(xy)}$, then $|f(xy)|$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module