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MIT 18.01 Single Variable Calculus (OCW)
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dy/dx=2x/5y y=1 x=0
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\(\large\color{slate}{ \displaystyle \frac{dy}{dx}=\frac{2x}{5y} }\) if `y=1` and `x=0` \(\large\color{slate}{ \displaystyle \frac{dy}{dx}=\frac{2(0)}{5(1)}=0 }\)
5y dy=2x dx integrating \[5\int\limits y~dy=2 \int\limits x~dx\] \[5\frac{ y^2 }{ 2 }=2 \frac{ x^2 }{ 2 }+c\] when x=0,y=1 \[\frac{ 5 }{ 2 }=0+c,c=\frac{ 5 }{ 2 }\] \[5\frac{ y^2 }{ 2 }=x^2+\frac{ 5 }{ 2 }\] \[5y^2=2 x^2+5\]
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