# Give the form of a particular solution of y^{(4)} + 5 y'' - 36 y = -4 e^{2 x} - 3 cos (3 x) - 2.

## Question:

Give the form of a particular solution of {eq}\displaystyle y^{(4)} + 5 y'' - 36 y = -4 e^{2 x} - 3 \cos (3 x) - 2 {/eq}.

## Method of Undetermined Coefficients:

Method of Undetermined Coefficients is the one of the best methods to find the form of the particular solution of the non-homogeneous DE. We first find the terms in the complimentary function, which is a solution to the corresponding homogeneous equation. Then, we consider only those terms in the particular solution which can be obtained by differentiation of the given function on the right-hand side. If any term is duplicated, then that term is multiplied by {eq}x^m, {/eq} where m is the multiplicity of the root in the auxiliary equation.

a) The given differential equation is linear second-order non-homogeneous with constant coefficients.

{eq}\displaystyle y^{(4)} + 5 y'' - 36 y = -4 e^{2 x} - 3 \cos (3 x) - 2 {/eq}

For complementary function, we find the solution to the corresponding homogeneous differential equation which is given by {eq}y^{(4)} + 5 y'' - 36 y=0 \\ {/eq}

The auxiliary equation is {eq}m^{4}+5m^2-36=0\\ \Rightarrow m^{4}+9m^2-4m^2-36=0\\ \Rightarrow m^2(m^2+9)-4(m^2+9)=0\\ \Rightarrow (m^2-4)(m^2+9)=0\\ \Rightarrow (m+2)(m-2)(m^2+9)=0\\ {/eq} Therefore, {eq}m=\pm 2, \ \pm 3i. \\ {/eq}

Hence, the complementary function is {eq}\displaystyle CF=C_{1}e^{-2x}+C_{2}e^{2x}+C_3 \cos (3x)+C_4 \sin (3x). \\ {/eq}

Now, the right-hand side of the given equation is {eq}\displaystyle -4 e^{2 x} - 3 \cos (3 x) - 2. {/eq}

Its derivatives contain the terms {eq}\displaystyle e^{2 x}, \ \cos (3x), \ \sin (3x), \ 1. {/eq}

Now, {eq}\displaystyle e^{2x} {/eq} is a solution in CF corresponding to m=2 which has multiplicity 1, we multiply the duplicated derivative term by '{eq}x {/eq}'.

Similarly, {eq}\displaystyle \cos (3x), \ \sin (3x) {/eq} are solutions in CF corresponding to {eq}m=\pm 3i {/eq} which has multiplicity 1, we multiply the duplicated derivative term by '{eq}x {/eq}'.

Therefore, the form of particular solution is given by:

{eq}\displaystyle y_p(x)=\color{blue}{ Axe^{2 x}+Bx \cos (3x)+Cx \sin (3x)+D}. {/eq}