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Question

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = 1$. First,we find th

Vedclass · Accepted Answer

$y = 1 + c{e^{-x}}$

Vedclass · Answer

(A) The given differential equation is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = 1$ and $Q = 1$. First,we find the Integrating Factor $(I.F.)$: $I.F. = e^{\int P dx} = e^{\int 1 dx} = e^x$. The general solution is given by $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + c$. Substituting the values: $y \cdot e^x = \int 1 \cdot e^x dx + c$. Integrating the right side: $y \cdot e^x = e^x + c$. Dividing both sides by $e^x$: $y = \frac{e^x}{e^x} + \frac{c}{e^x} = 1 + c{e^{-x}}$. Thus,the correct option is $A$.

The solution of the differential equation $\frac{dy}{dx} + y = 1$ is

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