Question

How can I solve this differential equation $y''' - 6y'' = 3 - \cos x$ by undetermined coefficients?

Answer 1

In this question we have to solve the differential equation using undetermined coefficients, first solve the characteristic equation of this differential, and use the fact that homogeneous part of solution is ${y_c} = {C_1} + {C_2}x + {C_3}{e^{6x}}$ ,as the terms ${C_2}x$ and ${C_1}$ have part of homogenous solution, particular solution must be in form: ${y_p} = A{x^2} + B\cos x + C\sin x$ , now deriving the ${y_p}$ three times and comparing this with the given equation we will get the coefficients, and by adding the equations ${y_c}$ and ${y_p}$ , we will get the required answer.

Complete step by step solution:
Given equation is $y''' - 6y'' = 3 - \cos x$ ,
First the characteristic equation of this differential one will be,
$\Rightarrow {x^3} - 6{x^2} = 0$ ,
Now taking common term we get,
$\Rightarrow {x^2}\left( {x - 6} \right) = 0$ ,
Now this can be written as,
$\Rightarrow x \cdot x\left( {x - 6} \right) = 0$ ,
Hence the roots are,
$\Rightarrow$ ${x_1} = 0$ , ${x_2} = 0$ and ${x_3} = 6$ ,
We know that homogeneous part of solution is ${y_c} = {C_1} + {C_2}x + {C_3}{e^{6x}}$ ,as the terms ${C_2}x$ and ${C_1}$ have part of homogenous solution, particular solution must be in form: ${y_p} = A{x^2} + B\cos x + C\sin x$ ,
Now differentiating both sides we get,
$\Rightarrow y{'_p} = 2Ax - B\sin x + C\cos x$ ,
Now again differentiating on both sides we get,
$\Rightarrow y'{'_p} = 2A - B\cos x - C\sin x$ ,
Now again differentiating on both sides we get,
$\Rightarrow y''{'_p} = B\sin x - C\cos x$ ,
From the above we get, by substituting the values in the given equation we get,’
$\Rightarrow$ $y''' - 6y'' = B\sin x - C\cos x - 6\left( {2A - B\cos x - C\sin x} \right)$ ,
Now simplifying we get,
$y''' - 6y'' = B\sin x - C\cos x - 12A + 6B\cos x + 6C\sin x$ ,
Now combing the like terms we get,
$\Rightarrow y''' - 6y'' = \sin x\left( {B + 6C} \right) + \cos x\left( {6B - C} \right) - 12A$ ,
Now we know that $y''' - 6y'' = 3 - \cos x$ , now comparing the terms we get,
$\Rightarrow 3 - \cos x = \sin x\left( {B + 6C} \right) + \cos x\left( {6B - C} \right) - 12A$ ,
Now comparing we get,
$\Rightarrow - 12A = 3$ , $6B - C = - 1$ and $B + 6C = 0$ ,
Now simplifying we get,
$\Rightarrow A = \dfrac{{ - 3}}{{12}} = \dfrac{{ - 1}}{4}$ ,
Now solving the equations $6B - C = - 1 - - - - (1)$ and
$B + 6C = 0 - - - - (2)$ , we get,
$\Rightarrow$ $B = - 6C$ , now substituting the value in (1) we get,
$\Rightarrow 6\left( { - 6C} \right) - C = - 1$ ,
Now simplifying we get,
$\Rightarrow - 36C - C = - 1$ ,
Now adding we get,
$\Rightarrow - 37C = - 1$ ,
Now taking out negative sign we get,
$\Rightarrow 37C = 1$ ,
Now dividing 37 on both sides we get,
$\Rightarrow \dfrac{{37C}}{{37}} = \dfrac{1}{{37}}$ ,
Now simplifying we get,
$\Rightarrow C = \dfrac{1}{{37}}$ ,
Now substituting the values in the equation ${y_p} = A{x^2} + B\cos x + C\sin x$ , we get,
$\Rightarrow {y_p} = \dfrac{{ - 1}}{4}{x^2} - \dfrac{6}{{37}}\cos x + \dfrac{1}{{37}}\sin x$ ,
So, finally required equation will be sum of ${y_c}$ and ${y_p}$ , so, equation is,
$\Rightarrow {y_c} + {y_p} = {C_1} + {C_2}x + {C_3}{e^{6x}}\dfrac{{ - 1}}{4}{x^2} - \dfrac{6}{{37}}\cos x + \dfrac{1}{{37}}\sin x$ ,

By solving differential equation $y''' - 6y'' = 3 - \cos x$ by undetermined coefficients is equal to ${y_c} + {y_p} = {C_1} + {C_2}x + {C_3}{e^{6x}}\dfrac{{ - 1}}{4}{x^2} - \dfrac{6}{{37}}\cos x + \dfrac{1}{{37}}\sin x$ .

Note: The method of finding a particular solution ${y_p}$ , first make an assumption about the form of ${y_p}$ , then find the coefficients, and general rule for assumption is that linear combination of all the linearly independent types of functions that aside from repeated differentiations. No function in the assumed ${y_p}$ duplicates any part of ${y_c}$ .