The solution of {e^{dy/dx}} = (x + 1),y(0) = | vedclass

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

Question

(B) Given the differential equation: ${e^{dy/dx}} = (x + 1)$.Taking the natural logarithm on both sides: $\frac{dy}{dx} = \log(x + 1)$.Separating the

Vedclass · Accepted Answer

$y = (x + 1)\log |x + 1| - x + 3$

Vedclass · Answer

(B) Given the differential equation: ${e^{dy/dx}} = (x + 1)$.Taking the natural logarithm on both sides: $\frac{dy}{dx} = \log(x + 1)$.Separating the variables: $dy = \log(x + 1) dx$.Integrating both sides: $y = \int \log(x + 1) dx$.Using integration by parts,let $u = \log(x + 1)$ and $dv = dx$. Then $du = \frac{1}{x+1} dx$ and $v = x + 1$.$y = (x + 1)\log(x + 1) - \int \frac{x+1}{x+1} dx$.$y = (x + 1)\log(x + 1) - \int 1 dx$.$y = (x + 1)\log(x + 1) - x + C$.Given the initial condition $y(0) = 3$,substitute $x = 0$ and $y = 3$:$3 = (0 + 1)\log(0 + 1) - 0 + C$.$3 = 1 \cdot \log(1) + C$.Since $\log(1) = 0$,we get $C = 3$.Therefore,the solution is $y = (x + 1)\log |x + 1| - x + 3$.

The solution of ${e^{dy/dx}} = (x + 1)$,$y(0) = 3$ is

Similar Questions

Solve the differential equation $\frac{dy}{dx} = \frac{y(1+x)}{-x(1+y)}$.

The solution of $\frac{dy}{dx} = (x+y)^2$ is

The solution of the differential equation $2x \frac{dy}{dx} - y = 3$ represents a family of

The general solution of the differential equation $\frac{d y}{d x}+\frac{y^2+y+1}{x^2+x+1}=0$ is

The general solution of the differential equation $y \log y \, dx - x \, dy = 0$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module