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form the DE of family of circles touching y-axis at origin .
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\[x^2+y^2=2ax \] After D. w.r.t.x \[2x+2yy'=2a\]
Now if i Differentiate it again im getting a different answer
different answre from?
I have the solution
Can you do this problem ?
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I am trying to understand what it's asking to do
okay
\[x^2+y^2=2ax\] was ^ this given or you came up with it?
I came up with it
circle \[x^2+y^2=a^2\] we need to move it to right by a/2 \[(x-a/2)^2+y^2=a^2\] \[x^2-ax+a^2/4+y^2=a^2\]
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diff --> \[2x-a+2y y'=0\]
2x+2y y'= a
now ?
ill do it in diff eqn
ok :)
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2y y' =-2x +a y'= -x/y + a/2y y'= (-2x+a)/2y
Given Answer is 2xyy' + x^2 = y^2
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