If y=y(x) is the solution of the differential | vedclass

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

Question

(B) Given the differential equation: $\frac{5+e^x}{2+y} \frac{dy}{dx} = -e^x$. Separating the variables,we get: $\int \frac{dy}{2+y} = \int \frac{-e^x

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(B) Given the differential equation: $\frac{5+e^x}{2+y} \frac{dy}{dx} = -e^x$. Separating the variables,we get: $\int \frac{dy}{2+y} = \int \frac{-e^x}{e^x+5} dx$. Integrating both sides,we obtain: $\ln(y+2) = -\ln(e^x+5) + C_1$. This simplifies to: $\ln(y+2) + \ln(e^x+5) = C_1$,which implies $(y+2)(e^x+5) = C$. Given the condition $y(0)=1$,we substitute $x=0$ and $y=1$: $(1+2)(e^0+5) = C \Rightarrow 3(1+5) = C \Rightarrow C = 18$. Thus,the equation is $(y+2)(e^x+5) = 18$,or $y+2 = \frac{18}{e^x+5}$. To find $y(\ln 13)$,substitute $x = \ln 13$: $y+2 = \frac{18}{e^{\ln 13}+5} = \frac{18}{13+5} = \frac{18}{18} = 1$. Therefore,$y = 1 - 2 = -1$.

If $y=y(x)$ is the solution of the differential equation $\frac{5+e^x}{2+y} \cdot \frac{dy}{dx} + e^x = 0$ satisfying $y(0)=1$,then a value of $y(\log_e 13)$ is

Similar Questions

A short bar magnet placed with its axis at $30^{\circ}$ with an external field of $800$ $Gauss$, experiences a torque of $0.016 ext{ N.m}$. The work done in moving it from most stable to most unstable position is $\alpha \times 10^{-3} ext{ J}$. The value of $\alpha$ is . . . . . . . . . .

If $y$ is a function of $x$ and $\log(x+y)=2xy$,then the value of $y'(0)$ is

Product $(A)$ is

In a series of $2n$ observations,half of the observations are equal to $a$ and the remaining are equal to $-a$. If the standard deviation of the observations is $2$,then what is the value of $|a|$?

Which of the following elements forms a stable $+4$ oxidation state?

Vedclass Test Series

Exam Paper Generator

Online Exam Module