dy/dx - y/(x+1) = x y(1)= 0, x>0 integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)] y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1) 0= 1/2 -(ln2)/2 + C/3 y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1) is the answer right?
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Originally Posted by twilightstr dy/dx - y/(x+1) = x y(1)= 0, x>0 integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)] y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1) 0= 1/2 -(ln2)/2 + C/3 y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1) is the answer right? You can always check your answer by substituting it into the differential equation and seeing if it works.
Originally Posted by twilightstr dy/dx - y/(x+1) = x y(1)= 0, x>0 integral of [Ae^-ln(x+1)(x)]/[Ae^-ln(x+1)] y= x/(x+1) - ln(x+1)/(x+1) + c/(x+1) 0= 1/2 -(ln2)/2 + C/3 y= x/(x+1) -ln(x+1)/(x+1) + (-3+ln2)/2(x+1) is the answer right? No. The integrating factor is . So multiply both sides of the equation by and you get . Can you go from here?
It is even easier to see that that y does not satisfy y(1)= 0!
thanks but do i have the answer now: y= x^2-xln(x+1)+(-1 + ln2)x +x-ln(x+1)-1+ln2
Originally Posted by twilightstr thanks but do i have the answer now: y= x^2-xln(x+1)+(-1 + ln2)x +x-ln(x+1)-1+ln2 Let's see... . We're told so . So .
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