- #1
Pengwuino
Gold Member
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Ok I have a problem here. I have gotten to the unfortunate point where I feel like I'm nearly done, Mathematica has given me a solution that is correct… but of course, I can't figure out how I was suppose to get from point A to B. I currently have:
2x - 6y\sqrt {x^2 + 1} \frac{{dy}}{{dx}} = 0,y(0) = 3
So I did this…
\begin{array}{l}<br /> 2xdx = 6y(x^2 + 1)^{1/2} \frac{{dy}}{{dx}} \\ <br /> \int {2x(x^2 + 1)^{ - 1/2} dx = } \int {6ydy} \\ <br /> u = x^2 + 1 \\ <br /> dx = \frac{{du}}{{2x}} \\ <br /> 2u^{1/2} = 3y^2 + c \\ <br /> 2(x^2 + 1)^{1/2} = 3y^2 + c \\<br /> \end{array}
Kinda not sure where to go from here…. Or if I did that right in the first place
The correct answer according to Mathematica is...
\frac{{\sqrt {{\rm 25 + 2}\sqrt {{\rm 1 + x}^{\rm 2} } } }}{{\sqrt 3 }}
So yup, put x=0 and 3 pops out.
2x - 6y\sqrt {x^2 + 1} \frac{{dy}}{{dx}} = 0,y(0) = 3
So I did this…
\begin{array}{l}<br /> 2xdx = 6y(x^2 + 1)^{1/2} \frac{{dy}}{{dx}} \\ <br /> \int {2x(x^2 + 1)^{ - 1/2} dx = } \int {6ydy} \\ <br /> u = x^2 + 1 \\ <br /> dx = \frac{{du}}{{2x}} \\ <br /> 2u^{1/2} = 3y^2 + c \\ <br /> 2(x^2 + 1)^{1/2} = 3y^2 + c \\<br /> \end{array}
Kinda not sure where to go from here…. Or if I did that right in the first place
The correct answer according to Mathematica is...
\frac{{\sqrt {{\rm 25 + 2}\sqrt {{\rm 1 + x}^{\rm 2} } } }}{{\sqrt 3 }}
So yup, put x=0 and 3 pops out.
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