Solve by using the method of undetermined coefficients. 𝑥 2 𝑑 2𝑦 𝑑𝑥 2 − 3𝑥 𝑑𝑦 𝑑𝑥 + 5𝑦 = 𝑥 2 sin 𝑙𝑜𝑔𝑥

Dear Student

Given Differential equation is𝑥²𝑦″−3𝑥𝑦′+5𝑦=𝑥²𝑠𝑖𝑛(𝑙𝑜𝑔𝑥).
x²y″−3xy′+5y=x2sin(logx).
Since it is second order linear differential equation (Euler - Cauchy equation),
the homogeneous part of equation can be solved as follows,

Let𝑦=𝑥𝑟,y=xr,to get the characteristic equation, which is

𝑟(𝑟−1)−3𝑟+5=𝑟²−4𝑟+5=0r(r−1)−3r+5
=r²−4r+5=0

𝑟=2+𝑖,2−𝑖.r=2+i,2−i.

Thus the homogeneous solution𝑦ℎ=𝑐1𝑥²𝑠𝑖𝑛(𝑙𝑜𝑔𝑥)+𝑐2𝑥²𝑐𝑜𝑠(𝑙𝑜𝑔𝑥).yh=c1x²sin(logx)+c2x²cos(logx).

Now the particular solution is found by many ways, here I will just guess the particular solution to be𝑎𝑥2𝑙𝑛(𝑥)𝑐𝑜𝑠(𝑙𝑜𝑔𝑥).ax2ln(x)cos(logx).

To find the value of the constant𝑎,a,we substitute the the particular solution into differential equation.

𝑎𝑥²((𝑙𝑜𝑔𝑥+3)𝑐𝑜𝑠(𝑙𝑜𝑔𝑥)−(3𝑙𝑜𝑔𝑥+2)𝑠𝑖𝑛(𝑙𝑜𝑔𝑥))+𝑎𝑥²((2𝑙𝑜𝑔(𝑥)+1)𝑐𝑜𝑠(𝑙𝑜𝑔𝑥)−𝑙𝑜𝑔𝑥𝑠𝑖𝑛(𝑙𝑜𝑔𝑥))−4𝑎𝑥²𝑙𝑛𝑥𝑐𝑜𝑠(𝑙𝑜𝑔𝑥)=−2𝑎𝑥²𝑠𝑖𝑛(𝑙𝑜𝑔𝑥)=𝑥²𝑠𝑖𝑛(𝑙𝑜𝑔𝑥)ax²((logx+3)cos(logx)−(3logx+2)sin(logx))+ax²((2log(x)+1)cos(logx)−logxsin(logx))−4ax²lnxcos(logx)=−2ax²sin(logx)=x²sin(logx)

So𝑎=−12.a=−12.

Thus the solution is𝑦=𝑐1𝑥²𝑠𝑖𝑛(𝑙𝑜𝑔𝑥)+𝑐2𝑥²𝑐𝑜𝑠(𝑙𝑜𝑔𝑥)−12𝑥²𝑙𝑛(𝑥)𝑐𝑜𝑠(𝑙𝑜𝑔𝑥).

Thanks