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\[y(x)=2+\int\limits_{1}^{x}\left(\begin{matrix}dt \\ ty(t)\end{matrix}\right)\]
x>0
Did you try differentiating ?
\[\frac{ dy }{ dx }=\frac{ dx }{ xy(x) }\]
Is this part correct?
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dx wont be there after differentiating
\[\frac{ dy }{ dx }=\frac{ 1 }{ xy }\]
separate variables and solve the DE
\[\frac{ y^2 }{ 2 }=\ln \left| x \right|+C\]
\[C=\frac{ y^2 }{ 2 }-\ln \left| x \right|\]
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But what do I do from here?
use the intial condition to find C
y(1) = 2
plugin x = 1, y = 2 in the final equation
\[C=\frac{ 2^2 }{ 2 }-\ln \left| 1 \right|\]
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So C=2
\[y=\sqrt{2\ln \left| x \right|+4}\]
Is this right?
Looks good!
Thank you!
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