What is a solution to the differential equation #dy/dx=-x/y# with the particular solution #y(1)=-sqrt2#?

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Answer 1 · 2016-09-25T18:31:45

Multiply both sides by #ydx#,integrate, and solve for the initial conditions

#ydy = -xdx#

#(1/2)y^2 = -1/2x^2 + C#

Because the form of C is arbitrary, we can write the above as our friend the circle:

#x^2 + y^2 = C^2#

Forcing the initial condition:

#1^2 + (-sqrt2)^2 = C^2#

#C^2 = 3#

The equation becomes:

#x^2 + y^2 = (sqrt3)^2#