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The solution of (dy)/(dx)+y=e^(-x), y(0)...

The solution of `(dy)/(dx)+y=e^(-x), y(0)=0 " is"`

A

`y=e^(-x)(x-1)`

B

`y=xe^(x)`

C

`y=xe^(-x)+1`

D

`y=xe^(-x)`

Text Solution

Verified by Experts

The correct Answer is:
D
Given that, `" "(dy)/(dx)+y=e^(-x)`
which is a linear differential equation
Here, `P=1 and Q=e^(-x)`
`" "IF=e^(intdx)=e^(x)`
The general solution is
`" "y*e^(x)=inte^(-x)*e^(x)dx+C`
`rArr" "ye^(x)=intdx+C`
`rArr" "ye^(x)=x+C" "`...(i)
When x=0 and y=0 then, 0=0+C `rArr` C=0
Eq. (i) becomes `y*e^(x)=xrArr=xe^(-x)`
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