# Use reduction of order, with y(x) = u(x) cos(3x) to find the general solution to the following ODE:  y'' + 9y = sin(3x)Thanks

Luca B. | Certified Educator

calendarEducator since 2011

starTop subjects are Math, Science, and Business

You need to solve the differential equation, hence, you need to find the particular solution and the complementary solution.

You need to determine the particular solution using the method of undetermined coefficients.

You should consider a particular solution to the given differential equation such taht:

`y = x(a sin 3x + b cos 3x)`

You need to differentiate y with respect to x such that:

`y' = (dy)/(dx) = a sin 3x + b cos 3x + x(3a cos 3x - 3b sin 3x)`

You need to differentiate y' with respect to x such that:

`y'' = 3a cos 3x - 3b sin 3x + 3a cos 3x - 3b sin 3x + x(-9a sin 3x - 9bcos 3x)`

`y'' = 6a cos 3x - 6b sin 3x - 9ax sin 3x - 9bx cos 3x`

You should substitute `6a cos 3x - 6b sin 3x - 9ax sin 3x - 9bx cos 3x `  for `y''` in differential equation such that:

`6a cos 3x - 6b sin 3x - 9ax sin 3x - 9bx cos 3x + 9xa sin 3x + 9xb cos 3x = sin3x`

Reducing like terms yields:

`6a cos 3x - 6b sin 3x = sin 3x`

Equating the coefficients of sin 3x and cos 3x of both sies yields:

`6a = 0 => a = 0`

`-6b = 1 => b = -1/6`

Hence, evaluating the particular solution of differential equation yields `y = -(1/6)x cos 3x` .

You need to find the complementary solution solving the following equation such that:

`y'' + 9y = 0`

You should substitute `e^(cx)`  for y such that:

`(e^(cx))'' + 9e^(cx) = 0`

`c^2*e^(cx) + 9e^(cx) = 0`

Factoring out `e^(cx)`  yields:

`e^(cx)(c^2 + 9) = 0`

Since `e^(cx)>0`  for any c, hence `c^2 + 9 = 0`  such that:

`c^2 + 9 = 0 => c^2 = -9 => c = +-sqrt(-9)`

Using complex number theory, `i^2 = -1` , yields:

`c = +-3i`

Hence, evaluating the complementary solutions to the differential equation yields `y_1 = c_1*e^(3i*x), y_2 = c_2*e^(-3i*x)`

Using Euler's identity yields:

`y = y_1 + y_2`

`y = c_1(cos 3x + i*sin 3x) + c_2(cos 3x- i*sin 3x)`

`y = cos 3x*(c_1 + c_2) + i*sin 3x(c_1 - c_2)`

Using the following constants yields:

`c = c_1+ c_2 ; d = i*(c_1-c_2)`

`y = c cos 3x + d sin 3x`

You need to evaluate the general solution to the given differential equation such that:

`y = c cos 3x + d sin 3x - (1/6)x cos 3x`

Hence, evaluating the solution of second order differential equation yields `y = c cos 3x + d sin 3x - (1/6)x cos 3x.`

check Approved by eNotes Editorial