Differential Equations 1
Quiz 4 Solutions
Solve the differential equation: dy / dx = (y + sqrt(xy) ) / x, x, y, > 0.
This was intended as an exercise in solving homogeneous equations, and is fairly easy if you take that route.
Two students noticed that the technique we used for Bernoulli equations works, and one of those students was successful in carrying out all of the steps.
A few students tried the substitution y = v 2x and that could have been successful.
Solution 1 (Homogeneous Equation).
dy / dx = y / x + sqrt(y/x), therefore the substitution y = vx will work. Calculate dy / dx = v + x dv /dx and substitute, to get:
v + x dv /dx = v + sqrt(v) ). Thus, x dv /dx =sqrt(v), and then dv / sqrt(v) = dx / x. Thus,
∫ dv / sqrt(v) = ∫ dx / x,
2 sqrt(v) =ln(x) + c, (note that x > 0) and v = ( (1/2) ln(x) + k) 2. This gives,
y =xv = x ( (1/2) ln(x) + k) 2.
Solution 2 (Bernoulli Equation).
Write the differential equation as dy / dx - (1 / x ) y = x -1/2y 1/2.
This fits the form for the Bernoulli equation,
dy / dx +p(x) y = g(x) y n, with n = 1/2.
Thus, the tranformation v = y 1-n = y 1/2
will tranform the equation into a linear first order differential equation.
It may be easier to write y as a function of v; that is,
y = v 2. Then, dy / dx = 2 v dv / dx. Substituting into the equation in Bernoulli form, we obtain:
2 v dv / dx - (1/x) v 2 = x -1/2v, or, simplifying,
dv / dx - ( 1/ (2x) ) v = (1/2) x -1/2.
This is a linear equation, and has integrating factor μ = exp( ∫ - dx / (2x) ) = x -1/2,
ignoring the constant of integration, as we do for the integrating factor. The transformed equation is:
x -1/2 dv / dx - ( 1/2 ) x -3/2 v = 1 / (2x) .
d( x -1/2v ) = dx / (2x) .
∫ d( x -1/2v ) = ∫ dx / (2x) .
x -1/2v = (1/2) ln(x) +c . (Again, x > 0, and thus ln|x| = ln(x).)
x -1/2y 1/2 = (1/2) ln(x) +c .
x -1y = ( (1/2) ln(x) +c ) 2.
y = x ( (1/2) ln(x) +c ) 2.
Solution 3 (Ad Hoc): Try the substitution y = w 2x for yourself.
URL http//math.fau.edu/locke/courses/DiffEqns/Q4Sols.htm