# Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients) Find the

Second Order Nonhomogeneous Differential Equation (Method of Undetermined Coefficients)
Find the general solution of the following Differential equation $y{}^{″}-2{y}^{\prime }+10y={e}^{x}\mathrm{cos}\left(3x\right)$. We know that the general solution for 2nd order Nonhomogeneous differential equations is the sum of ${y}_{p}+{y}_{c}$ where ${y}_{c}$ is the general solution of the homogeneous equation and ${y}_{p}$ the solution of the nonhomogeneous. Therefore ${y}_{c}={e}^{x}\left({c}_{1}\mathrm{cos}\left(3x\right)+{c}_{2}\mathrm{cos}\left(3x\right)\right)$. Now we have to find ${y}_{p}$. I know in fact that ${y}_{c}={e}^{x}\left({c}_{1}\mathrm{cos}\left(3x\right)+{c}_{2}\mathrm{cos}\left(3x\right)\right)$. Now we have to find yp. I know in fact that ${y}_{p}={e}^{×}\frac{\mathrm{sin}\left(3x\right)}{6}$ but i do not know how to get there.
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j3jell5
Step 1
since the RHS is a soln of the homogeneous eqn, we can try x multiplied by it. let ${y}_{p}$ be a linear combination of $x{e}^{x}\mathrm{cos}\left(3x\right)$ and $x{e}^{x}\mathrm{sin}\left(3x\right)$.
${y}_{p}=ax{e}^{x}\mathrm{cos}\left(3x\right)+bx{e}^{x}\mathrm{sin}\left(3x\right)$
Step 2
Let $E={D}^{2}-2D+10$, where D is the derivative operator.
$E\left[{y}_{p}\right]=-6a{e}^{x}\mathrm{sin}\left(3x\right)+6b{e}^{x}\mathrm{cos}\left(3x\right)$
$E\left[{y}_{p}\right]={e}^{x}\mathrm{cos}\left(3x\right)$
equating coeffs,
${y}_{p}=\frac{1}{6}x{e}^{x}\mathrm{sin}\left(3x\right)$
###### Not exactly what you’re looking for?
Bethany Mills
Step 1
After a lot of trial and error i finnaly understand the strategy behind this type of differential equation. So here's is what i did in order to find the solution (I will provide as much information as i can) In order to calculate a nonhomogeneous differntial equation we must first find the general solution of the homogeneous. So $y{}^{″}-2{y}^{\prime }+10y=o$
(1) using the Auxiliary equation we can easily find that the general solution is ${y}_{h}={e}^{x}\left({C}_{1}\mathrm{cos}\left(3x\right)+{C}_{2}\mathrm{sin}\left(3x\right)$. Now as cineel mentioned above we can try ${y}_{h}$ multiplied by x. Thus ${y}_{p}=Ax{e}^{x}\mathrm{cos}\left(3x\right)+Bx{e}^{x}\mathrm{sin}\left(3x\right)$. Now we have to differentiate ${y}_{p}$ two times.
So ${y}_{p}=Ax{e}^{x}\mathrm{cos}\left(3x\right)+Bx{e}^{x}\mathrm{sin}\left(3x\right)$ (1)
${y}_{p}^{\prime }=A\left[{e}^{x}\mathrm{cos}\left(3x\right)+x{e}^{x}\mathrm{cos}\left(3x\right)-x{e}^{x}3\mathrm{sin}\left(3x\right)\right]+B\left[{e}^{x}\mathrm{sin}\left(3x\right)+x{e}^{x}\mathrm{sin}\left(3x\right)+3x{e}^{x}\mathrm{cos}\left(3x\right)$ (2)
${y}_{p}{}^{″}=A\left[-8x{e}^{x}\mathrm{cos}\left(3x\right)-6{e}^{×}\mathrm{sin}\left(3x\right)+2{e}^{x}\mathrm{cos}\left(3x\right)-3{e}^{x}\mathrm{sin}\left(3x\right)\right]+B\left[-8x{e}^{x}\mathrm{sin}\left(3x\right)+6{e}^{×}\mathrm{cos}\left(3x\right)+2{e}^{x}\mathrm{sin}\left(3x\right)+6{e}^{x}\mathrm{cos}\left(3x\right)\right]$ (3)
Step 2
Therefore substituting (1)(2)(3) in our original differential equation we end up with $-3A{e}^{x}\mathrm{sin}\left(3x\right)+6B{e}^{x}\mathrm{cos}\left(3x\right)={e}^{x}\mathrm{cos}\left(3x\right)$
So
Thus (1) Now becomes ${y}_{p}=\frac{1}{6}x{e}^{x}\mathrm{sin}\left(3x\right)$
So ${y}_{p}+{y}_{h}={e}^{x}\left[{C}_{1}\mathrm{cos}\left(3x\right)+{C}_{2}\mathrm{sin}\left(3x\right)\right]+\frac{1}{6}{e}^{×}\mathrm{sin}\left(3x\right)$