Consider the differential equation frac{dy}{d | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{dx} = \frac{1}{ax + 4y + 7}$.Taking the reciprocal,we get: $\frac{dx}{dy} = ax + 4y + 7$.Rearranging t

Vedclass · Accepted Answer

Only $B$ and $A$ are true

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \frac{1}{ax + 4y + 7}$.Taking the reciprocal,we get: $\frac{dx}{dy} = ax + 4y + 7$.Rearranging the terms,we have: $\frac{dx}{dy} - ax = 4y + 7$.This is a linear differential equation of the form $\frac{dx}{dy} + Px = Q$,where $P = -a$ and $Q = 4y + 7$.Since the equation can be written in the form $\frac{dx}{dy} + Px = Q$,it is linear in $x$.Because the equation is linear in $x$,it is not linear in $y$.Thus,statement $A$ is true and statement $B$ is true.The integrating factor $(IF)$ is given by $e^{\int P dy} = e^{\int -a dy} = e^{-ay}$.Therefore,statement $D$ is false,and statement $C$ is false.Hence,only $A$ and $B$ are true.

Similar Questions

Let $x = x(y)$ be the solution of the differential equation $y^2 dx + (x - \frac{1}{y}) dy = 0$. If $x(1) = 1$,then $x(\frac{1}{2})$ is:

Let $f$ be a non-negative function defined on $\left[0, \frac{\pi}{2}\right]$. If $\int_0^x \left(f^{\prime}(t)-\sin 2t\right) dt = \int_x^0 f(t) \tan t dt$ and $f(0)=1$,then find $\int_0^{\frac{\pi}{2}} f(x) dx$.

Let $y(x)$ be the solution of the differential equation $(x \log x) \frac{dy}{dx} + y = 2x \log x$,$(x \ge 1)$. Then $y(e)$ is equal to: $[y(1) = 0]$

The solution of the differential equation $(1 + y^2) + (x - e^{\tan^{-1}y}) \frac{dy}{dx} = 0$ is

If $y(x)$ satisfies the differential equation $y' + y = 2(\sin x + \cos x)$ and $y(0) = 1$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module