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

(D) Given the differential equation: $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$.This can be rewritten as $e^{y} \frac{dy}{dx} - e^{y} = e^{x}$.Let $e^{

Vedclass · Accepted Answer

$1+\log _{e} 2$

Vedclass · Answer

(D) Given the differential equation: $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$.This can be rewritten as $e^{y} \frac{dy}{dx} - e^{y} = e^{x}$.Let $e^{y} = t$. Then,differentiating with respect to $x$,we get $e^{y} \frac{dy}{dx} = \frac{dt}{dx}$.Substituting these into the equation,we get the linear differential equation: $\frac{dt}{dx} - t = e^{x}$.The integrating factor ($I$.$F$.) is $e^{\int -1 dx} = e^{-x}$.Multiplying both sides by the $I$.$F$.,we get $\frac{d}{dx}(t e^{-x}) = e^{x} \cdot e^{-x} = 1$.Integrating both sides with respect to $x$,we get $t e^{-x} = x + c$.Substituting $t = e^{y}$,we have $e^{y} e^{-x} = x + c$,which simplifies to $e^{y-x} = x + c$.Given $y(0) = 0$,we substitute $x = 0$ and $y = 0$: $e^{0-0} = 0 + c \Rightarrow 1 = c$.So,the solution is $e^{y-x} = x + 1$.To find $y(1)$,substitute $x = 1$: $e^{y-1} = 1 + 1 = 2$.Taking the natural logarithm on both sides,$y - 1 = \log_{e} 2$.Therefore,$y(1) = 1 + \log_{e} 2$.

If $y=y(x)$ is the solution of the differential equation $e^{y}\left(\frac{dy}{dx}-1\right)=e^{x}$ such that $y(0)=0,$ then $y(1)$ is equal to

Similar Questions

For $x \in R$,let the function $y(x)$ be the solution of the differential equation $\frac{dy}{dx} + 12y = \cos \left(\frac{\pi}{12} x\right)$ with $y(0) = 0$. Then,which of the following statements is/are $TRUE$?

Which of the following equations is a linear differential equation?

The solution of $(x+y+1) \frac{dy}{dx} = 1$ is

The general solution of a differential equation of the type $\frac{dx}{dy} + P_{1}x = Q_{1}$ is

If $y(t)$ is a solution of $(1 + t)\frac{dy}{dt} - ty = 1$ and $y(0) = -1$,then $y(1)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module