Answer to Question #219351 in Differential Equations for Umra

Question

Solve the following (Euler-Cauchy) differential equations:
(x^4D^4+6x^3D^3+9x^2D^2+3xD+1)y=(1+lnx)^2

Answer 1

By the method of Euler-Cauchy , we have to solve this differential equation -

$=(x^{4}D^{4}+6x^{3}D^{3}+9x^{2}D^{2}+3xD+1)y=(1+lnx)^{2}$

Now let , $x=e^{t}$ , Taking log both the side of given equation , we get ,

$=lnx=t$

Now we know that $,$ the terms on left hand side of given differential equation can be reduces into linear differential equation , by putting the formulas -

$=x^{4}D^{4}=D(D-1)(D-2)(D-3)y$

$=x^{3}D^{3}=$ $D(D-1)(D-2)y$

$=x^{2}D^{2}=D(D-1)y$

$=xD=Dy$

Putting all these above values in the given equations , we get -

$=$ $(D(D-1)(D-2)(D-3)+6D(D-1)(D-2)+9D(D-1)+3D+1)y=(1+t)^{2}$

Now , solving we get as -

$=$ $(D^{4}+2D^{2}+1)y=(1+t)^{2}$

It's Auxiliary equation will be -

$=m^{4}+2m^{2}+1=0$

$=m^{4}+m^{2}+m^{2}+1=0$

$=m^{2}(m^{2}+1)+1(m^{2}+1)=0$

$=(m^{2}+1)^{2}=0$

$m={\pm}i$

so $CF=[(c_1+c_2x)cosx+(c_3+c_4x)sinx]$

Now to find $PI$ of given function , we get -

$PI=\dfrac{1}{D^{4}+2D^{2}+1}$ $(1+t)^{2}$

$PI=(1+D^{4}+2D^{2})^{-1}(1+t)^{2}$

Now using Binomial expansion of $(1+x)^{-1}=1-x+x^{2}+....$

Now applying this series in above equation we get ,

$PI=(1-D^{4}-2D^{2})(1+2t+t^{2})$

$PI=1+2t+t^{2}-4=t^{2}+2t-3$

Solution of given Partial differential equation can be written as -

$Y=CF+PI=$ $[(c_1+c_2x)cosx+(c_3+c_4x)sinx]+(lnx)^{2}+2lnx-3$

which is required solution of given partial differential equation .