find the general solution of (x^2)y"+5xy'+3y=lnx for x>0
Answers
Answer:
Euler Equations
In this section, we will investigate the solutions of the most simple type of differential equations with regular singular points
x2y'' + axy' + by = 0
We can immediately see that 0 is a regular singular point of the differential equation since
xp(x) = a and x2q(x) = b
To solve the differential equation we assume that a solution is of the form
y = xr
Now we proceed in a similar way to how we tackled second order differential equations with constant coefficients. We take two derivatives.
y' = r xr-1 y'' = r (r - 1) xr-2
Next plug these into the original differential equation
x2 r (r - 1)xr-2 + ax r xr-1 + b xr = 0
r (r - 1)xr + ar xr + b xr = 0 Multiplying the exponents
r (r - 1) + ar + b = 0 Dividing by xr
r2 + (a - 1) r + b = 0
We define
F(r) = r2 + (a - 1) r + b
This is a quadratic in r. We will see as with second order homogeneous linear differential equations with constant coefficients, there will be three cases.
Example (Real Distinct Roots)
Solve
x2y'' + 5xy' + 3y = 0
Solution
Let
y = xr
We take two derivatives.
y' = r xr-1 y'' = r (r - 1) xr-2
Next plug these into the original differential equation
x2 r (r - 1)xr-2 + 5x r xr-1 + 3xr = 0
r (r - 1)xr + 5r xr + 3xr = 0 Multiplying the exponents
r (r - 1) + 5r + 3 = 0 Dividing by xr
r2 + 4 r + 3 = 0
(r + 3)(r + 1) = 0
r = -3 or r = -1
The general solution is
y = c1x -3 + c2x -1
Example (Repeated Roots)
Solve
x2y'' + 7xy' + 9y = 0
Solution
Let
y = xr
We have
F(r) = r2 + 6r + 9 = (r + 3)2
which has the repeated root
r = -3
Hence a solution is
y1 = x -3
This gives us one solution and we could get another solution by reduction of order. However, there is a more clever way.
We notice that both F(r) and F'(r) are zero at r = -3.
Notice also that the partial derivative
(xr)r = xr ln x
We have
Lr(xr) = [xrF(r)]r
or
L(xr ln r) = F(r) xr ln x + xr Fr(r) = (r + 3)2 xr ln x +2xr (r + 3)
Now plug in r = -3 to get
L(x -3 ln x) = 0
Hence
y2 = x -3 ln x
The general solution is
y = c1x -3 + c2x -3 ln x
Example (Complex Roots)
Solve
x2y'' + 5xy' + 8y = 0
Solution
Let
y = xr
We have
F(r) = r2 + 4r + 8
which has complex roots
r = 2 + 4i and r = 2 - 4i
We get the solutions
y1 = x2 + 4i and y2 = x2 - 4i
As with constant coefficients, we would like to express the solution without complex numbers. We have
x -2 + 2i = e(-2 + 2i)ln x = x-2e2lnx i = x-2[cos(2ln x) + i sin(2ln x)]
Similarly
x -2 - 2i = e(-2 - 2i)ln x = x -2e-2lnx i = x-2[cos(2ln x) - i sin(2ln x)]
By playing with constants we get the two solutions
y1 = x -2 cos(2ln x) and y2 = x -2sin(2ln x)
The general solution is
y = c1x -2 cos(2ln x) + c2x -2 sin(2ln x) = x -2[c1cos(2ln x) + c2sin(2ln x)]
In summary, we have
Theorem: Solutions to Euler Equations
Let
x2y'' + axy' + by = 0
and let
F(r) = r2 + (a - 1)r + b
have roots r1 and r2
Case 1: If r1 and r2 are real and distinct, then the general solution is
y = c1xr1 + c2xr2
Case 2: If r1 = r2 = r then the general solution is
y = c1xr + c2xr ln x
Case 3: If r1 = l + mi and r2 = l - mi then the general solution is
y = xl (c1 cos(m lnx) + c2 sin(m lnx))
Answer:
y
Solve the differential equation x^ 2 y^ prime prime +5xy^ prime +3y=0