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Consider the differential equation xv dv dx + v 2 = 32x 2 . (a) Solve the differential equation using an appropriate integrating factor.?
- az_lenderLv 72 months ago
Your typography is weird. Did you mean
xv dv/dx + v^2 = 32x^2 ?
First I'll solve it using v = xu, dv/dx = u + x du/dx.
ux^2 (u + x du/dx) + x^2u^2 = 32x^2,
so wherever x is not 0, we get
u(u + x du/dx) + u^2 = 32 =>
u^2 + xu du/dx + u^2 = 32 =>
xu du/dx = 32 - u^2 =>
u du/(32 - u^2) = dx/x =>
-(1/2)ln(32 - u^2) = ln(x) + C1 =>
1/sqrt(32 - u^2) = cx =>
1/sqrt(32 - v^2/x^2) = cx, or
x*sqrt(32 - v^2/x^2) = C.
Can I go back and solve the equation with an integrating factor? It's not coming to me right now...