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The form of the particular solution of the fourth-order, linear, nonhomogeneous ODE \begin{equation} e^{2x}y^{(4)}+64e^{2x}y+1=x\cos{(2x)} \end{equation} is found. There are standard forms for the particular solutions of nonhomogeneous equations with constant coefficients. The first step is to transform this equation to an equation with constant coefficients by dividing both side by $e^{2x}$. This gives after a bit or rearrangement \begin{equation} y^{(4)}+64y=\left[x\cos{(2x)}-1\right]e^{-2x} \end{equation} The nonhomogeneous term of the transformed equation is $\left[x\cos{(2x)}-1\right]e^{-2x}$. Before proceeding to the determination of the particular solution, one must check that none of the terms within the nonhomogeneous term solve the corresponding homogeneous equation which is called the complimentary solution. The characteristic equation associated with the homogeneous equation is $r^4+64=0$ which has the repeated roots $r=\pm i2\sqrt{2}$ each with multiplicity two. Thus the linearly independent solution of the homogenous equations are \begin{align} y_1(x)=&\cos{\left(2\sqrt{2}x\right)}&y_2(x)=&\sin{\left(2\sqrt{2}x\right)}\\ y_3(x)=&x\cos{\left(2\sqrt{2}x\right)}&y_4(x)=&x\sin{\left(2\sqrt{2}x\right)}. \end{align} Clearly, none of the terms within the nonhomogeneous term solve the corresponding homogeneous equation. It follows that the form of the particular solution is \begin{equation} y_p(x)=\left[\left(Ax+B\right)\left(C\cos{(2x)}+D\sin{(2x)}\right)+E\right]e^{-2x} \end{equation} You can see a third-order nonhomogeneous solution at the link https://tutorial.math.lamar.edu/classes/de/HOUndeterminedCoeff.aspx
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Thu, 10 Dec 2020 18:41 GMT