# find the general solution of the following differential equation.2x^2 d^2y/dx^2+5x dy/dx + y = x^2 -x

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### 1 Answer

The solution to `2x^2(d^2y)/(dx^2)+5x(dx)/(dy)+y=0` is found by substituting t=ln(x) and then solving the equation

`2(d^2y)/(d^2t)+3(dy)/(dt)+y=0`

We solve the characteristic equation `2gamma^2+5gamma+1=0` to get

`gamma=(-3+-sqrt(9-8))/4=(-3+-sqrt(1))/4` gives gamma=-1 or -1/2

Gives the solutions `y=C_1e^(-t)+C_2e^(-t/2)=C_1/x + C_2/sqrt(x)`

Suppose the specific solution to the equation is y=Ax^2+Bx. `(dy)/(dx)=2Ax+B` and `(d^2y)/(dx^2)=2A` . Substituting we get

`2x^2(2A)+5x(2Ax+B)+(Ax^2+Bx)=x^2-x`

`4Ax^2+10Ax^2+5Bx+Ax^2+Bx=x^2-x`

`(4A+10A+A)x^2+(5B+B)x=x^2-x`

15A=1 and 6B=-1 so A=1/15, B=-1/6 So our solution is

` y=1/15x^2 - 1/6x + C_1/x + C_2/sqrt(x)`