Observe the following statements:A. Integrati | vedclass

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

Question

(C) For statement $A$:The given differential equation is $\frac{dy}{dx} + y = x^2$.Comparing this with the standard linear form $\frac{dy}{dx} + P(x)y

Vedclass · Accepted Answer

$A$ is true,$R$ is true,$R \Rightarrow A$

Vedclass · Answer

(C) For statement $A$:The given differential equation is $\frac{dy}{dx} + y = x^2$.Comparing this with the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$,we get $P(x) = 1$.The integrating factor $(IF)$ is given by $e^{\int P(x) dx} = e^{\int 1 dx} = e^x$.Thus,statement $A$ is true.For statement $R$:The standard form of a linear differential equation is $\frac{dy}{dx} + P(x)y = Q(x)$.The integrating factor is defined as $e^{\int P(x) dx}$.Thus,statement $R$ is true.Since statement $A$ is derived directly from the formula given in statement $R$,$R \Rightarrow A$ is true.Therefore,both statements are true and $R$ is the correct explanation for $A$.

Observe the following statements:
$A$. Integrating factor of $\frac{dy}{dx} + y = x^2$ is $e^x$.
$R$. Integrating factor of $\frac{dy}{dx} + P(x)y = Q(x)$ is $e^{\int P(x) dx}$.
Then,the true statement among the following is:

Similar Questions

The general solution of the differential equation $\frac{dy}{dx} + \frac{\sec x}{\cos x + \sin x} y = \frac{\cos x}{1 + \tan x}$ is

To reduce the differential equation $\frac{dy}{dx} + P(x)y = Q(x)y^n$ to the linear form,the substitution is

The general solution of the equation $\frac{dy}{dx} + \frac{1}{x}y = \frac{1}{x}e^x$ is

Let $y(x)$ be a solution of the differential equation $(1+e^x) y^{\prime}+y e^x=1$. If $y(0)=2$,then which of the following statements is (are) true?
$(A)$ $y(-4)=0$
$(B)$ $y(-2)=0$
$(C)$ $y(x)$ has a critical point in the interval $(-1,0)$
$(D)$ $y(x)$ has no critical point in the interval $(-1,0)$

An integrating factor of the differential equation $\frac{dy}{dx} + \frac{2xy}{1 - x^2} = \frac{x}{\sqrt{1 - x^2}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Observe the following statements:$A$. Integrating factor of $\frac{dy}{dx} + y = x^2$ is $e^x$.$R$. Integrating factor of $\frac{dy}{dx} + P(x)y = Q(x)$ is $e^{\int P(x) dx}$.Then,the true statement among the following is: