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Mathematics 13 Online
OpenStudy (anonymous):

differential equations

OpenStudy (anonymous):

OpenStudy (anonymous):

That is a separable equation \[ \frac {dy}{dx}= -y^2\\ -\frac {dy}{y^2} = dx\\ \frac 1 y = x + C\\ y= \frac {1}{x+C} \] The rest is easy

OpenStudy (anonymous):

y=0 is a solution that does not belong the family\( y=\frac 1 {x+C}\)

OpenStudy (anonymous):

For the last part \[ y(0)=\frac 1 { 0+C}= \frac 1 2\implies C=2\\ y=\frac 1 {x+ 2} \] is the answer to the last part.

OpenStudy (anonymous):

for a.) y=x^-1, y'=-1/x^2=-y^2 so y<=0? isit coz its negative and 1/(large no.) is actually 0?

OpenStudy (anonymous):

thanks for the other parts i appreciate! (:

OpenStudy (anonymous):

1/K is not actually equal to a zero for large K. Its limit is zero, but the there is no finite number K so that 1/K=0

OpenStudy (anonymous):

i see, got it okay thanks!(:

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