y''+4y'+4y=(e^(-2x)) / (x^2), x>0 Determine a particular solution of the nonhomogeneous differential equation using the method of variation of parameters.

Expert Answers

An illustration of the letter 'A' in a speech bubbles

`lambda^2+4lambda+4=0`

Solution is `(lambda+2)^2 =0`

`lambda = -2`

We then obtain `u_1=e^(-2x)` and `u_2 = xe^(-2x)`

The Wronskian is

`W=-e^(-2x)e^(-2x)(2x-1)+2xe^(-4x)=e^(-4x)`

Since the Wronksian is non zero the two functions are independent and therefore the two functions are the general solution to the differential equation.

We find the particular solution by...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

`lambda^2+4lambda+4=0`

Solution is `(lambda+2)^2 =0`

`lambda = -2`

We then obtain `u_1=e^(-2x)` and `u_2 = xe^(-2x)`

The Wronskian is

`W=-e^(-2x)e^(-2x)(2x-1)+2xe^(-4x)=e^(-4x)`

Since the Wronksian is non zero the two functions are independent and therefore the two functions are the general solution to the differential equation.

We find the particular solution by computing `A(x)u_1+B(x)u_2`

`A(x)=-int 1/Wu_2(x)b(x) dx = -int 1/e^(-4x) xe^(-2x) e^(-2x)/x^2 dx=-int 1/x dx`

So `A(x) = -ln(x)+C_1`

`B(x) = int 1/W u_1(x)b(x) dx = int 1/e^(-4x)e^(-2x)e^(-2x)/x^2 dx = int 1/x^2 dx`

So `B(x)=-1/x + C_2`

So our general solution is

`y = (-ln(x)+C_1)e^(-2x)+(-1/x+C_2)xe^(-2x)` Simplifying

`y = -(ln(x)+1)e^(-2x) + C_1e^(-2x) + C_2xe^(-2x)`

Approved by eNotes Editorial Team