The solution of the given differential equati | vedclass

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Question

(A) Given differential equation is $\frac{dy}{dx} + 2xy = y$. Rearranging the terms,we get $\frac{dy}{dx} + (2x - 1)y = 0$. This is a linear different

Vedclass · Accepted Answer

$y = ce^{x - x^2}$

Vedclass · Answer

(A) Given differential equation is $\frac{dy}{dx} + 2xy = y$. Rearranging the terms,we get $\frac{dy}{dx} + (2x - 1)y = 0$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 2x - 1$ and $Q = 0$. The integrating factor $(I.F.)$ is given by $I.F. = e^{\int P dx} = e^{\int (2x - 1) dx} = e^{x^2 - x}$. The general solution is $y 	imes (I.F.) = \int (Q 	imes I.F.) dx + c$. Substituting the values,$y 	imes e^{x^2 - x} = \int (0 	imes e^{x^2 - x}) dx + c$. $y 	imes e^{x^2 - x} = 0 + c$. $y = c e^{-(x^2 - x)} = c e^{x - x^2}$. Thus,the correct option is $A$.

The solution of the given differential equation $\frac{dy}{dx} + 2xy = y$ is

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