Inactive Tutor answered 06/27/13
Tutor
New to Wyzant
y^2/2=x^2+c
Plug in y(0) = 2,
2^2 / 2 = 0^2 + c
c = 2
Answer: y = sqrt[2(x^2 + 2)]
Sun K.
asked 06/27/13Solve the initial value problem: y'=2x/y, y(0)=2.
dy/dx=2x/y
y dy=2x dx
y^2/2=x^2+c
Now what?
Inactive Tutor answered 06/27/13
y^2/2=x^2+c
Plug in y(0) = 2,
2^2 / 2 = 0^2 + c
c = 2
Answer: y = sqrt[2(x^2 + 2)]
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