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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.?

### 1 Answer

- 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...