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  • Solution of dy/dx - y =1 is given by A. xy = -ex B. xy = -e-x C. xy = -1 D. Y = 2 ex – 1

Solution of \frac{d y}{d x}-y=1, y(0)=1 is given by
A.xy = -e^x

B. xy = -e^{-x}
C.xy = -1
D. y = 2 e^x- 1

Answers (1)

\\ \frac{d y}{d x}-y=1$ \\$\Rightarrow \frac{d y}{d x}=1+y$ \\$\Rightarrow \frac{\mathrm{dy}}{1+\mathrm{y}}=\mathrm{dx}$
Integrate
\\\Rightarrow \int \frac{d y}{1+y}=\int d x\\$ $\log (1+x)=x+c$
now it is given that y(0)=1 which means when  x=0, y=1 hence substitute x=0 and y=0 in (a)
\\\log (1+y)=x+c$ \\$\log (1+1)=0+c$ \\$\mathrm{c}=\log 2$
put \mathrm{c}=\log 2$  back in (a)

\\\log (1+y)=x+\log 2$ \\$\log (1+y)-\log 2=x$
using \log a-\log b=\log a / b$
\\\Rightarrow \log \frac{1+y}{2}=x$ \\$\Rightarrow \frac{1+y}{2}=e^{x}$ \\$1+y=2 e^{x}$ \\$y=2 e^{x}-1$
Hence solution of differential equation is \mathrm{y}=2 \mathrm{e}^{\mathrm{x}}-1$

Option D is correct.

Posted by

infoexpert22

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