# Solution of $\Large \frac{dy}{dx}$$-y=1,y(0)=1 is given by $(A)\;xy=-e^x\quad(B)\;xy=-e^{-x}\quad(C)\;xy=-1\quad(D)\;y=2e^x-1$ ## 1 Answer Toolbox: • A linear differential equation of the form \large\frac{dy}{dx}$$+Py=Q$ has a general solution $y e^{\large\int pdx}=\int Q.e^{\large \int pdx}.dx+c$. where $e^{\large \int pdx}$ is the integrating factor (I.F)
Given $\large\frac{dy}{dx}$$-y=1 This is a linear differential equation of the forn \large\frac{dy}{dx}$$+Py=Q$ where $P=-1$ and $Q=1$
Hence the solution is $y e^{\large\int pdx}=\int Q.e^{\large \int pdx}+c$.
Where $e^{\large\int pdx}$ is the integrating factor I.F
$\int pdx=\int -dx=-x$
Hence $I.F=e^{-x}$
Hence the solution is $ye^{-x}=\int e ^{-x}+c$
On integrating we get
$y e^{-x}=-e^{-x}+c$
It is given $y(0)=1$
This implies when $x=0,y=1$
Now substituting the values we get the value of c as
$ye^0=-e^0+c$
$(But \;e^0=1)$
$1=-1+c=>c=2$
Hence the required solution is $ye^{-x}=e^{-x}+2$
Multiplying throughout by $e^{+x}$ we get
$y=-1+2e^{x}$
or $y=2e^x-1$
Hence the correct option is $D$