Observe the following statements:I. If dy+2xy | vedclass

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

Question

(A) $I$. Given $dy+2xy dx=2e^{-x^2} dx$.Dividing by $dx$,we get $\frac{dy}{dx}+2xy=2e^{-x^2}$.This is a linear differential equation of the form $\fra

Vedclass · Accepted Answer

Both $I$ and $II$ are true

Vedclass · Answer

(A) $I$. Given $dy+2xy dx=2e^{-x^2} dx$.Dividing by $dx$,we get $\frac{dy}{dx}+2xy=2e^{-x^2}$.This is a linear differential equation of the form $\frac{dy}{dx}+Py=Q$,where $P=2x$ and $Q=2e^{-x^2}$.Integrating factor $I.F. = e^{\int P dx} = e^{\int 2x dx} = e^{x^2}$.The general solution is $y(I.F.) = \int Q(I.F.) dx + c$.$ye^{x^2} = \int 2e^{-x^2} \cdot e^{x^2} dx + c = \int 2 dx + c = 2x+c$.Thus,statement $I$ is true.$II$. Given $ye^{x^2}-2x=c$.Differentiating with respect to $x$: $\frac{d}{dx}(ye^{x^2}) - \frac{d}{dx}(2x) = 0$.$y(2x e^{x^2}) + e^{x^2} \frac{dy}{dx} - 2 = 0$.$e^{x^2} \frac{dy}{dx} = 2 - 2xye^{x^2}$.$\frac{dy}{dx} = 2e^{-x^2} - 2xy$.Therefore,$dx = \frac{dy}{2e^{-x^2}-2xy}$.Thus,statement $II$ is true.

Observe the following statements:
$I$. If $dy+2xy dx=2e^{-x^2} dx$,then $ye^{x^2}=2x+c$
$II$. If $ye^{x^2}-2x=c$,then $dx=\frac{dy}{2e^{-x^2}-2xy}$
Which of the following is a correct statement?

Similar Questions

Find the general solution of the differential equation: $x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x$.

The equation of the curve passing through the origin and satisfying the equation $(1 + {x^2})\frac{{dy}}{{dx}} + 2xy = 4{x^2}$ is

Let $y=y(x)$ be the solution of the differential equation $x^{4}dy + (4x^{3}y + 2\sin x)dx = 0$, $x>0$, $y(\frac{\pi}{2})=0$. Then $\pi^{4}y(\frac{\pi}{3})$ is equal to:

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Observe the following statements:$I$. If $dy+2xy dx=2e^{-x^2} dx$,then $ye^{x^2}=2x+c$$II$. If $ye^{x^2}-2x=c$,then $dx=\frac{dy}{2e^{-x^2}-2xy}$Which of the following is a correct statement?