anonymous
  • anonymous
Solve the given initial value problem for y =f(x) dy/dx = x+1/x^1/2
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
y = 5, x=4
anonymous
  • anonymous
so it does look to me like you can separate the variables to dy=x^(-1/2)(x+1)dx. Then, go ahead and distribute the x^(-1/2) so dy = x^(1/2) + x^(-1/2)dx. Then integrate. After you integrate you will have that c value, so let y=5, and sub in 4 for your x values, then you solve for C. I got y=(2/3)x^(3/2)+2x^(1/2)-(25/3).
anonymous
  • anonymous
What do you mean by "After you integrate you will have that c value"?? I'm teaching myself integrals, can you explain that a little bit please? Haha

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anonymous
  • anonymous
whenever you have an indefinite integral, your result is the equation of the antiderivative, + and arbitrary constant, so say with this eq. y=(2/3)x^(3/2)+2x^(1/2) + C after integrating it. That is a reason why they use the initial value when solving differential equations, so that way you can find that value of C and have a real equation, instead of an arbitrary formula for one, which is what you end up with, with indefinite integrals.
anonymous
  • anonymous
So I take y=(2/3)x^(3/2)+2x^(1/2) + C and substitute in my x and y values? Then solve for C?
anonymous
  • anonymous
right. Know also that if you're teaching yourself integration, that unless it is a definite integral you always have that extra value.
anonymous
  • anonymous
I really appreciate the help, thank you!
anonymous
  • anonymous
no problem, good luck with your studies

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