Recent content by EmmanuelD
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Calculate the molar volume using the van der Waals equation.
Homework Statement Calculate the molar volume using the van der Waals equation of a gas at P=3000psia, T=60F. The critical pressure and temperature, Pc and Tc, are Pc=408psia, Tc=504F. Homework Equations The given van der Waals equations(s): (P+a/Vm^2)(Vm-b)=RT ----(1)...- EmmanuelD
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- Van der waals Volume
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- Forum: Biology and Chemistry Homework Help
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Bernoulli - Find the general solution
GOT IT! Sorry, I don't know if there's an option for deleting a topic. My apologies. Thanks, either way :)- EmmanuelD
- Post #2
- Forum: Calculus and Beyond Homework Help
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Bernoulli - Find the general solution
Bernoulli -- Find the general solution Homework Statement Find the general solution of: y'+xy=xy^3 Homework Equations Bernoulli's Equation The Attempt at a Solution y'+xy=xy^3 (y^-3)y'+x(y^-2)=x Let v=(y^-2), thus v'=((-2y^-3)y' Then, -v'/2+xv=x Multiply...- EmmanuelD
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- Bernoulli General General solution
- Replies: 2
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
Grr! Yes, thank you! I notice I make many silly mistakes when I rush through. So, it's ...+30y = C :] Thanks a BUNCH! I know it might be simplistic but our professor stressed how easy it was and just did one example (without the constant). So, not so bad after all. Again, thank...- EmmanuelD
- Post #11
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
Hehe, yes, I did in fact forget :) I reposted with my answer. y'=-x/5(y-3) (y-3) dy = -x/5 dx y2/2 - 3y = -x^2/10 +C 10*(x^2/10 + y^2/2 - 3y) = C Ans: x^2 + 5y^2 - 30y = C (?)- EmmanuelD
- Post #9
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
So, I got: x^2+5y^2-30y=C If you solved the problem, is this correct? Thanks!- EmmanuelD
- Post #8
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
Oops! I forgot what I was doing here. Back to business: y'=-1/y' Don't lose hope on me, I'm just burned out. I'm going to try this on paper. Thanks for the help thus far! :)- EmmanuelD
- Post #6
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
Or, taking a different route (inspired by your initial reply); C=(y-3)/x^5 Sub'd into y' y'=5x^4((y-3)/x^5)) y'=5/x(y-3) 1/(y-3) dy = 5/x dx ln(y-3) = 5*ln(x) + C y-3=x^5 + C ...but this still doesn't look right to me; unless I'm missing something.- EmmanuelD
- Post #5
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
I apologize; I just edited the post. Given: y=C*x^5 - 3 Solution for C: C=y/x^5 + 3 Then, as shown above, I sub'd in the value for C in the y': y'=5x^4(y/x^5 + 3) y'=5y/x + 15x^4 Now, I can't factor out a "y" to make this D.E. separable.- EmmanuelD
- Post #3
- Forum: Calculus and Beyond Homework Help
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Find the Orthogonal Trajectories For The Family of Curves
Hello, forum! I'm a newbie here. I've been visiting this site for a while but just recently joined. Anyways, I was wondering if anyone could help with this problem. I can find the orthogonal trajectories, however, this one is killing me because there is a constant. Allow me to type it below...- EmmanuelD
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- Curves Orthogonal Trajectories
- Replies: 10
- Forum: Calculus and Beyond Homework Help