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

Answer 1

Paisley Johnson

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#

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Answer 2

Ezekiel Alexander

The particular solution to the given differential equation ( \frac{dy}{dx} = -\frac{x}{y} ) with the initial condition ( y(1) = -\sqrt{2} ) is ( y(x) = -\sqrt{x^2 + 1} ).

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Answer from HIX Tutor

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