Ask your own question, for FREE!
Mathematics 20 Online
OpenStudy (anonymous):

Solve the seperable differential equation for. \frac(dy)(dx) = \frac(1+x)(xy^(6)) ; \ \ x \gt 0 Use the following initial condition: y(1) = 5 .

zepdrix (zepdrix):

\[\Large \frac{dy}{dx}=\frac{1+x}{xy^6}, \qquad\qquad x\gt0, \qquad\qquad y(1)=5\]

zepdrix (zepdrix):

What part are you stuck on maria? :o Understand the separation portion of it?\[\Large y^6\;dy=\frac{1+x}{x}\;dx\]

OpenStudy (anonymous):

No I don't!!! I forgot ittt

zepdrix (zepdrix):

\[\Large \frac{dy}{dx}=\frac{1+x}{xy^6}\]So we want all of our `y and dy` on one side of the equation with all of our `x and dx` on the other side :o So we start by multiplying both sides by y^6.\[\Large y^6\frac{dy}{dx}=\frac{1+x}{x}\]Right? +_+

OpenStudy (anonymous):

so what do I have to plug in to get y=? because that part makes sense I just don't know how to use that information further

zepdrix (zepdrix):

You need to integrate which will get rid of the differentials. Have you done that part yet? :o\[\Large \int\limits y^6\;dy=\int\limits \frac{1+x}{x}\;dx\]

OpenStudy (anonymous):

1/7y^7=(ln|x|)+x

zepdrix (zepdrix):

\[\Large \frac{1}{7}y^7=\ln|x|+x+C\] Ok good :) Now we can use our initial condition solve for our unknown constant.

zepdrix (zepdrix):

\[\Large y(1)=5\]Plugging in gives us,\[\Large \frac{1}{7}5^7=\ln|1|+1+C\]

OpenStudy (anonymous):

11160.71=10.320155+c

zepdrix (zepdrix):

Nooo, no decimals! :O And what happened on the right side there..? Recall: ln1=0

OpenStudy (anonymous):

111.60.71=1+C

OpenStudy (anonymous):

11160=C

zepdrix (zepdrix):

fine fine fine :) have your sloppy decimals lol that looks good. So now we plug our c back in. and since we're able to, we should probably solve for y.

zepdrix (zepdrix):

\[\Large \frac{1}{7}y^7=\ln|x|+x+11160\]

zepdrix (zepdrix):

\[\Large y=?\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!