- #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:
[tex]2x - 6y\sqrt {x^2 + 1} \frac{{dy}}{{dx}} = 0,y(0) = 3[/tex]
So I did this…
[tex]\begin{array}{l}
2xdx = 6y(x^2 + 1)^{1/2} \frac{{dy}}{{dx}} \\
\int {2x(x^2 + 1)^{ - 1/2} dx = } \int {6ydy} \\
u = x^2 + 1 \\
dx = \frac{{du}}{{2x}} \\
2u^{1/2} = 3y^2 + c \\
2(x^2 + 1)^{1/2} = 3y^2 + c \\
\end{array}[/tex]
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...
[tex]\frac{{\sqrt {{\rm 25 + 2}\sqrt {{\rm 1 + x}^{\rm 2} } } }}{{\sqrt 3 }}[/tex]
So yup, put x=0 and 3 pops out.
[tex]2x - 6y\sqrt {x^2 + 1} \frac{{dy}}{{dx}} = 0,y(0) = 3[/tex]
So I did this…
[tex]\begin{array}{l}
2xdx = 6y(x^2 + 1)^{1/2} \frac{{dy}}{{dx}} \\
\int {2x(x^2 + 1)^{ - 1/2} dx = } \int {6ydy} \\
u = x^2 + 1 \\
dx = \frac{{du}}{{2x}} \\
2u^{1/2} = 3y^2 + c \\
2(x^2 + 1)^{1/2} = 3y^2 + c \\
\end{array}[/tex]
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...
[tex]\frac{{\sqrt {{\rm 25 + 2}\sqrt {{\rm 1 + x}^{\rm 2} } } }}{{\sqrt 3 }}[/tex]
So yup, put x=0 and 3 pops out.
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