The general solution of the differential equa | vedclass

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

Question

(D) Given the differential equation: $(y^2+x+1) dy = (y+1) dx$. Rearranging the terms to form a linear differential equation in $x$: $\frac{dx}{dy} =

Vedclass · Accepted Answer

$\frac{x+2}{y+1}+\log (y+1)^2=y+c$

Vedclass · Answer

(D) Given the differential equation: $(y^2+x+1) dy = (y+1) dx$. Rearranging the terms to form a linear differential equation in $x$: $\frac{dx}{dy} = \frac{y^2+x+1}{y+1} = \frac{y^2+1}{y+1} + \frac{x}{y+1}$. $\frac{dx}{dy} - \frac{1}{y+1} x = \frac{y^2+1}{y+1}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y+1}$ and $Q(y) = \frac{y^2+1}{y+1}$. The integrating factor $IF = e^{\int P(y) dy} = e^{-\int \frac{1}{y+1} dy} = e^{-\log(y+1)} = \frac{1}{y+1}$. The solution is $x \cdot IF = \int Q(y) \cdot IF dy + c$. $x \cdot \frac{1}{y+1} = \int \frac{y^2+1}{(y+1)^2} dy + c$. Using the substitution $y^2+1 = (y+1)^2 - 2y - 1 + 1 = (y+1)^2 - 2(y+1) + 2$: $\frac{x}{y+1} = \int \left( 1 - \frac{2}{y+1} + \frac{2}{(y+1)^2} \right) dy + c$. $\frac{x}{y+1} = y - 2 \log |y+1| - \frac{2}{y+1} + c$. Multiplying by $(y+1)$: $x = y(y+1) - 2(y+1) \log |y+1| - 2 + c(y+1)$. Alternatively,rearranging the integral result: $\frac{x+2}{y+1} = y - 2 \log |y+1| + c$. Thus,$\frac{x+2}{y+1} + 2 \log |y+1| = y + c$,which is $\frac{x+2}{y+1} + \log (y+1)^2 = y + c$.

The general solution of the differential equation $(y^2+x+1) dy = (y+1) dx$ is

Similar Questions

If the length of the sub-tangent at any point $P(x, y)$ on a curve $f(x, y) = 0$ is $x + 7y^2$, then $f(x, y) =$

If $y=y(x)$ is the solution of the differential equation $\frac{dy}{dx} + 2y \tan x = \sin x$ with the condition $y(\frac{\pi}{3}) = 0$,then the maximum value of the function $y(x)$ over $\mathbb{R}$ is equal to:

The solution of the differential equation $(x+1) \frac{dy}{dx} - xy = 1$,satisfying $y(0) = 1$ is

Find the general solution of the differential equation: $(1+x^{2}) dy + 2xy dx = \cot x dx$ where $x \neq 0$.

The general solution of the differential equation $\frac{dy}{dx} = y \tan x - y^2 \sec x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module