# Solving a second order non-homogeneous equation using undertermined coefficients-superposition approach: y'' + 4y = 3sin2x I get 0=3sin2x even if I change the right side to 6sinxcosx.

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### 1 Answer

The general solution to the given second order, non-homogeneous differential equation is `y = y_h + y_p` , where `y_h` represents the homogeneous solution to equation `y'' + 4y = 0` and `y_p` represents the particular solution that you need to find using the method of undetermined coefficients.

You need to solve first the homogeneous equation, such that:

`y'' + 4y = 0`

Considering `y = e^(kx)` , you may evaluate the second order derivative y'', such that:

`y' = k*e^(kx) => y'' = k^2*e^(kx)`

You need to replace `k^2*e^(kx)` for y'' and `e^(kx)` for y, such that:

`k^2*e^(kx) + 4e^(kx) = 0`

Factoring out `e^(kx)` yields:

`e^(kx)*(k^2 + 4) = 0 => k^2 + 4 = 0 => k = +-sqrt(-4) => k_(1,2) = +-2i`

Since `k_(1,2) ` are complex numbers, hence, the solution to the homogeneous differential equation is the following:

`y_h = e^(ax)(c_1*cos(bx) + c_2*sin(bx))`

`a = 0, b = 2`

`y_h = e^0*(c_1*cos(2x) + c_2*sin(2x))`

`y_h = c_1*cos(2x) + c_2*sin(2x)`

`c_1,c_2` are arbitrary constants

You need to evaluate the particular solution `y_p` using the method of undetermined coefficients, hence, you need to start by performing succesive differentiation of the function `y = 3sin 2x` , such that:

`y' = 6cos 2x `

`y'' = -12 sin 2x`

`y''' = -24 cos 2x`

`y^(iv) = 48 sin 2x`

Performing succesive differentiation yields the set of functions, such that: `{cos 2x, -sin 2x, -cos 2x, sin 2x}.`

Since the identified terms are not the terms of homogeneous solution, they may represent independent linear solution to the given equation, such that:

`y_p = acos 2x - bsin 2x - ccos 2x + dsin 2x`

You need to differentiate the solution with respect to x, such that:

`y'_p = -2asin 2x - 2bcos 2x + 2c sin 2x + 2d cos 2x`

`y''_p = -4a*cos 2x + 4bsin 2x + 4c cos 2x - 4d sin 2x`

Replacing this solution into the given original equation, yields:

`-4a*cos 2x + 4bsin 2x + 4c cos 2x - 4d sin 2x + 4(acos 2x - bsin 2x - ccos 2x + dsin 2x) = 3 sin 2x`

`-4a*cos 2x + 4bsin 2x + 4c cos 2x - 4d sin 2x + 4acos 2x - 4bsin 2x - 4ccos 2x + 4dsin 2x =3 sin 2x`

Factoring out `sin2x` and `cos 2x ` to the left side, yields:

`sin 2x*(4b - 4d - 4b + 4d) + cos 2x*(-4a + 4c + 4a - 4c) = 3 sin 2x`

`0 = 3 sin 2x` invalid

Since the particular solution cannot be determined, there exists a problem in the content provided by the problem (the content can be typed incorrectly).

**Hence, the general solution to the the given second order, non-homogeneous differential equation is `y =c_1*cos(2x) + c_2*sin(2x) + y_p` , where `y_p` cannot be determined.**