Please help me to solve it,its a differential equations y+x(dy/dx)=3 with (x=1,y=1) thx before....
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You should isolate the variables such that:
`x(dy/dx)=3-y => (dx)/x = (dy)/(3-y)`
Integrating both sides yields:
`int (dy)/(3-y) = int (dx)/x`
You should use substitution to evaluate `int (dy)/(3-y)` such that:
`3-y = t => -dy = dt`
`int (dy)/(3-y) = int -(dt)/t = -ln|t| + c`
`int (dy)/(3-y) = -ln|3-y| + c`
`-ln|3-y| = ln|x| + c`
You need to find the constant c, hence, you need to substitute 1 for x and 1 for y in `-ln|3-y| = ln|x| + c` such that:
`-ln(3-1) = ln1 + c`
Since `ln 1 = 0 => c = -ln 2 = ln(1/2)`
`-ln|3-y| = ln|x| + ln(1/2) => ln(1/(3-y)) = ln|x| + ln(1/2)`
Using logarithmic identities yields:
`ln(1/(3-y)) = ln (x/2) => 1/(3-y) = x/2 => 2 = x(3-y)`
`2 = 3x - xy => 2 - 3x = -xy => y = (3x-2)/x`
Hence, evaluating the solution to the given equation, under the given conditions, yields `y = (3x-2)/x` .
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