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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)` |
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Answer» 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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