How do you find the particular solution to #ysqrt(1-x^2)y'-xsqrt(1-y^2)=0# that satisfies y(0)=1?

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Answer 1 · 2017-03-07T10:18:47

#sqrt[1 -y^2]=sqrt[1 - x^2]-1#

Grouping variables

#(y dy)/sqrt[1 - y^2]=(x dx)/sqrt[1 - x^2]#

so it is a separable differential equation.

We have also

#int (x dx)/sqrt[1 - x^2] = -sqrt[1 - x^2]# and

#int (y dy)/sqrt[1 - y^2] = -sqrt[1 -y^2]# so

#-sqrt[1 -y^2]=-sqrt[1 - x^2]+C#

and #y(0)=1# so

#-sqrt[1 -1^2]=-1+C->C = 1#

so

#sqrt[1 -y^2]=sqrt[1 - x^2]-1#