Answer to Question #200430 in Differential Equations for akasha

Question

Solve the following by Cauchy Euler differential Equation 4x^2 (d^2 y)/(dx^2 )-4x dy/dx+3y=sin⁡(ln(-x))            x<0

Answer 1

Solution

Substitution t=ln(-x), y(x)=f(ln(-x))=f(t):

dy/dx = (df/dt)/x => xdy/dx = df/dt

d²y/dx² = (d²f/dt² - df/dt)/x² => x²d²y/dx² = d²f/dt² - df/dt  

4(d²f/dt² - df/dt)-4 df/dt+3 = sin(t) => d²f/dt² - 2df/dt+3f/4 = sin(t)/4

Characteristic equation

k²-2k+3/4=0 => k_1,2 = 1±0.5 => k₁ = 0.5, k₂ = 1.5

Solution of homogeneous equation f₀(t)=Ae^t/2+Be^3t/2  

Partial solution of nonhomogeneous equation f₁(t) = Csin(t)+Dcos(t)

-D-2C+3D/4=0  => 2C=-D/4 => D=-8C

-C+2D+3C/4=1/4 => -C+8D=1 => C=-1/65, D=8/65

f(t) = f₀(t)+ f₁(t) = Ae^t/2+Be^3t/2 –( sin(t)+8cos(t))/65 

Solution of given equation:

y(x) = f(t) = f(ln(-x)) = Ae^ln(-x)/2+Be^{3 ln(-x)/2} –(sin(ln(-x))+8cos(ln(-x)))/65 

y(x) = A(-x)^1/2+B(-x)^3/2-( sin(ln(-x))+8cos(ln(-x)))/65

Answer

y(x) = A(-x)^1/2+B(-x)^3/2-( sin(ln(-x))+8cos(ln(-x)))/65