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- anonymous

Solve the given initial value problem for y =f(x)
dy/dx = x+1/x^1/2

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- anonymous

Solve the given initial value problem for y =f(x)
dy/dx = x+1/x^1/2

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- anonymous

y = 5, x=4

- 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

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

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

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

right. Know also that if you're teaching yourself integration, that unless it is a definite integral you always have that extra value.

- anonymous

I really appreciate the help, thank you!

- anonymous

no problem, good luck with your studies

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