# Find the solution of the following diferential equation satisfying the giving initial conditions: y"+4y=x2 + 3ex , y(0)=0, y'(0)=2. The Method of Undetermined Coefficients I think left side is the...

Find the solution of the following diferential equation satisfying the giving initial conditions:

y"+4y=x2 + 3ex , y(0)=0, y'(0)=2.

The Method of Undetermined Coefficients

I think left side is the characteristic equation. So, y''+4y=0

r^2+4=0

r=+2, -2

The general solution is yg = C1cos2x + c2sin2x

but I am not sure how to get the particular

solution for this equation.

answer is y=(7/10sin2x) - (19/40cos2) + (1/4)x2 - (1/8) + (3/5)et

### 1 Answer | Add Yours

Given `y''+4y=x^2+3e^x;y(0)=0;y'(0)=2`

We have already found the general solution to be `y=c_1cos(2x)+c_2sin(2x)` , so we need to find the particular solution.

The particular solution will be the sum of the particular solutions to `(d^2y)/(dx^2)+4y=x^2` and `(d^2y)/(dx^2)+4y=3e^x` .

(1) The particular solution to the first equation is of the form `p_1=a_1+a_2x+a_3x^2` and the particular solution to the second equation is of the form `p_2=a_4e^x` .

So the particular solution is `p(x)=a_1+a_2x+a_3x^2+a_4e^x` .

(2) To solve for the `a_i"s"` compute:

`(d^2p(x))/(dx^2)=d^2/(dx^2)(a_1+a_2x+a_3x^2+a_4e^x)=2a_3+a_4e^x`

Then `(d^2p(x))/(dx^2)+4p(x)=x^2+3e^x`

`=>(2a_3+a_4e^x)+4(a_1+a_2x+a_3x^2+a_4e^x)=x^2+3e^x`

`=>(4a_1+2a_3)+4a_2x+4a_3x^2+5a_4e^x=x^2+3e^x`

Equating the coefficients yields:

`4a_1+2a_3=0`

`4a_3=1`

`5a_4=3`

`4a_2=0`

So `a_1=-1/8,a_2=0,a_3=1/4,a_4=3/5`

(3) Now `p(x)=(3e^x)/5+x^2/4-1/8` . The general solution is:

`y=(3e^x)/5+x^2/4-1/8+c_1cos(2x)+c_2sin(2x)`

To use the initial conditions we need `y'` :

`(dy)/(dx)=(3e^x)/5+x/2-2c_1sin(2x)+2c_2cos(2x)`

`y(0)=0 => 3/5-1/8+c_1cos(0)+c_2sin(0)=0`

`=>c_1+19/40=0 =>c_1=-19/40`

`y'(0)=2 => 3/5-2c_1sin(0)+2c_2cos(0)=2`

`=> 2c_2+3/5=2 => c_2=7/10`

(4) So the solution is:

`y=(3e^x)/5+x^2/4-1/8-19/40cos(2x)+7/10sin(2x)`