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Frustrating Bernoulli Equationby 1d20
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#1
Feb1612, 01:03 PM

P: 12

I've been unable to fully solve this: [itex]\frac{dy}{dx} + y = xy^4[/itex]
The U: [itex]u = y^{3}[/itex], so [itex]y = u^\frac{1}{3}[/itex], and [itex]\frac{dy}{dx} = \frac{1}{3}u^\frac{4}{3}\frac{du}{dx} [/itex] The substitution: [itex]\frac{1}{3}u^\frac{4}{3}\frac{du}{dx} + u^\frac{1}{3} = xu^\frac{4}{3}[/itex] Simplified: [itex]\frac{du}{dx}  3u = 3x[/itex] AKA: [itex]\frac{du}{dx} + 3x = 3u[/itex] I've tried solving the resulting equation as a linear: [itex]p(x) = 3x[/itex], so the integrating factor is [itex]e^{\frac{3}{2}x^2}[/itex]. Which creates this unworkable equation: [itex]e^{\frac{3}{2}x^2}\frac{du}{dx} + 3xe^{\frac{3}{2}x^2} = 3ue^{\frac{3}{2}x^2}[/itex] And I've tried solving it as a homogenous ([itex]3udx  3xdx  du = 0[/itex]): [itex]u = vx[/itex], so [itex]du = vdx + xdv[/itex] Subing those in creates this unworkable mess: [itex]3vxdx  3xdx  vdx  xdv = 0[/itex] Which leaves me stuck, because I've only learned to solve separable, exact, homogenous, linear and bernoullis. 


#2
Feb1612, 02:47 PM

P: 5

The equation is most easily solved at this stage: [itex]\frac{du}{dx}  3u = 3x[/itex]
The integrating factor is just [itex]e^{3x}[/itex]. Solve from there. 


#3
Feb1612, 02:49 PM

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[itex]\frac{du}{dx}  3u = 0[/itex] plus a particular integral of [itex]\frac{du}{dx}  3u = 3x[/itex] which is going to be of the form u(x) = a + bx, for some constants of a and b. 


#4
Feb1612, 02:53 PM

P: 5

Frustrating Bernoulli Equation
Both methods solve the equation in about 2 lines, but yeah you can do complimentary function and particular integral if you prefer.



#5
Feb1612, 02:55 PM

P: 5

Also the particular integral is u(x) = a  3x. You don't get two arbitrary constants from a first order ODE



#6
Feb1612, 02:57 PM

P: 5

Sorry it's a + x



#7
Feb1612, 04:58 PM

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#8
Feb1612, 05:53 PM

P: 12

Thanks for the help!
A final question: I thought all integrating factors must have the p(x) part integrated, but this seems to be an exception. How come? 


#9
Feb1612, 08:09 PM

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For a linear equation like this one, any solution of ##du/dx 3u = 3x## is the sum of the solutions of two equations, the "given" equation ##du/dx 3u = 3x## and the simpler equation ##du/dx  3u = 0## You can easily find the general solution of the equation with 0 on the right hand side, and that solution ##Ce^{3x}## contans an arbitrary constant. You know there is only one arbitrary constant in the complete solution, so you don't need another one. Instead of finding the general solution of ##du/dx 3u = 3x##, now we only need one particular solution. You can usually guess the form of a particular solution from the function on the right hand side, and then equate the coefficients of different terms in the equation to find the unknown values. In this case, we can guess that ##u = ax + b## is a solution for some values of a and b. Then ##du/dx = a##, and the substituting in the differential equation we have ##a 3(ax + b) = 3x## To satisfy that equation for all values of ##x##, the constant terms and the coefficients of ##x## must both cancel out. So we have ##a  3b = 0## and ##3ax = 3x## for all ##x## which gives ##a = 1## and ##b = 1/3## and the general solution is ##u = Ce^{3x} + x + 1/3##. That method might seem longwinded (or even sneaky) but the real benefit comes because most of the "interesting" differential equations in physics and engineering are second order not first order, and finding integrating factors for second order equations is hard. (And most of the longwindedness was explaining it, not actually doing it). Alternatively you can solve this equation using an integrating factor without any "trickery". ##du/dx 3u = 3x## The integrating factor is ##e^{3x}## ##e^{3x}du/dx  3u e^{3x} = 3xe^{3x}## ##d/dx (ue^{3x}) = 3xe^{3x}## ##ue^3x = 3\int x e^{3x}\,dx## And you need to integrate by parts to do the integral on the right hand side. 


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