Question
solve the differential equation dy/dx=y^2/x^3 for y=f(x) with condition y(1)=1.
Answers
Reiny
dy/dx=y^2/x^3
(1/y^2) dy = (1/x^3) dx
∫(1/y^2) dy = ∫(1/x^3) dx
∫ y^-2 dy = ∫ x^-3 dx
- y^-1 = -(1/2)x^-2 + c
-1/y = -1/(2x^2) + c
at (1,1)
-1 = -1/2 + c
c = -1/2
-1/y = -1/(2x^2) - 1/2
= (-1 - x^2)/(2x^2)
1/y = (1 + x^2)/(2x^2)
y = 2x^2/(1+x^2)
f(x) = 2x^2/(1+x^2)
check my algebra, I should have written it out on paper first, did it on-screen.
That is always dangerous.
(1/y^2) dy = (1/x^3) dx
∫(1/y^2) dy = ∫(1/x^3) dx
∫ y^-2 dy = ∫ x^-3 dx
- y^-1 = -(1/2)x^-2 + c
-1/y = -1/(2x^2) + c
at (1,1)
-1 = -1/2 + c
c = -1/2
-1/y = -1/(2x^2) - 1/2
= (-1 - x^2)/(2x^2)
1/y = (1 + x^2)/(2x^2)
y = 2x^2/(1+x^2)
f(x) = 2x^2/(1+x^2)
check my algebra, I should have written it out on paper first, did it on-screen.
That is always dangerous.
Ke$ha
Thank you!