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Find the solution of the differential equation that satisfies the given initial condition. dy/dx=x/y , y(0)=-1

Sagot :

LammettHash

Separating variables, we have

[tex]\dfrac{dy}{dx} = \dfrac xy \implies y\,dy = x\,dx[/tex]

Integrate both sides.

[tex]\displaystyle \int y\,dy = \int x\,dx[/tex]

[tex]\dfrac12 y^2 = \dfrac12 x^2 + C[/tex]

Given that [tex]y(0)=-1[/tex], we find

[tex]\dfrac12 (-1)^2 = \dfrac12 0^2 + C \implies C = \dfrac12[/tex]

Then the particular solution is

[tex]\dfrac12 y^2 = \dfrac12 x^2 + \dfrac12[/tex]

[tex]y^2 = x^2 + 1[/tex]

[tex]y = \pm\sqrt{x^2 + 1}[/tex]

and because [tex]y(0)=-1[/tex], we take the negative solution to accommodate this initial value.

[tex]\boxed{y(x) = -\sqrt{x^2+1}}[/tex]

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