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How can I solve this differential equation for quadratic resistance?
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[QUOTE="fayled, post: 4630681, member: 497826"] I'm solving a differential equation to do with quadratic resistance and it seems to be acting very strangely - I get the opposite sign of answer than I should. If anybody could have a quick look through that would be much appreciated. For a particle moving downward and taking positive upwards, I have mdv/dt=-mg+bv[SUP]2[/SUP] The terminal velocity comes out at v[SUB]l[/SUB]=-√(mg/b) (negative as it obviously has its terminal velocity downwards). Substituting this in gives dv/dt=-g(1-(v/v[SUB]l[/SUB])[SUP]2[/SUP]) Now make the substitution z=v/v[SUB]l[/SUB] so dv/dt=v[SUB]l[/SUB]dz/dt. Then we obtain v[SUB]l[/SUB]dz/dt=-g(1-z[SUP]2[/SUP]) Noting that 1/1-z[SUP]2[/SUP]=0.5[(1/1+z)+(1/1-z)] and separating variables we get ∫[(1/1+z)+(1/1-z)]dz=-2g/v[SUB]l[/SUB]∫dt Integrating each side then gives (the initial conditions are v=0 at t=0 so the constant of integration is zero) ln(1+z/1-z)=-2gt/v[SUB]l[/SUB]. Next let k=v[SUB]l[/SUB]/2g so that ln(1+z/1-z)=-t/k e[SUP]-t/k[/SUP]=1+z/1-z e[SUP]-t/k[/SUP]-ze[SUP]-t/k[/SUP]=1+z z(1+e[SUP]-t/k[/SUP])=e[SUP]-t/k[/SUP]-1 z=(e[SUP]-t/k[/SUP]-1)/(e[SUP]-t/k[/SUP]+1) Therefore v=v[SUB]l[/SUB][(e[SUP]-t/k[/SUP]-1)/(e[SUP]-t/k[/SUP]+1)] Now, the correct answer should be v=v[SUB]l[/SUB][(1-e[SUP]-t/k[/SUP])/(1+e[SUP]-t/k[/SUP])] i.e v=-v[SUB]l[/SUB][(e[SUP]-t/k[/SUP]-1)/(e[SUP]-t/k[/SUP]+1)] which must be right because as t→∞, v→v[SUB]l[/SUB] which by the definition of v[SUB]l[/SUB] is expected. I've spent a lot of time trying to work out where I'm going wrong and it's driving me crazy. Thanks in advance :) [/QUOTE]
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How can I solve this differential equation for quadratic resistance?
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