Question

In Problems 1–26 solve the given differential equation by undetermined coefficients.

${\mathit{y}}^{\mathbf{\left(}\mathbf{4}\mathbf{\right)}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{1}{\mathbf{)}}^{\mathbf{2}}$

Short answer

$y=C_1\cos x+C_2\sin x+C_{3}x\cos x+C_{4}x\sin x+x^2-2x-3$

Step-by-step solution

Form complementary function

Consider the differential equation$y^{\left(4\right)}+2y^{\text{'}\text{'}}+y=(x-1{)}^2$

The auxiliary equation corresponding to homogenous equation is:

From $m^4+2m^2+1=0$we find$m_1=m_2=i,m_3=m_4=-i$

So, the complementary function is

$y_c=C_1\cos x+C_2\sin x+C_{3}x\cos x+C_{4}x\sin x$

Particular solution

Let the particular solution be $y_p=A{x}^2+Bx+C$

$$\begin{gathered} y_p\text{'}=2Ax+B \\ {y}_p\text{'}\text{'}=2A \\ {y}_p\text{'}\text{'}\text{'}=y_p^{\left(4\right)}=0 \end{gathered}$$

Substituting these into the given differential equation yields

$4A+A{x}^2+Bx+C=x^2-2x+1$

On equating the coefficients of like terms, we get

$$\begin{gathered} A=1 \\ B=-2 \end{gathered}$$

And localid="1663909994312" $4A+C=1\Rightarrow C=-3$

So, $y_p=x^2-2x-3$

Thus, the general solution is

$y=y_c+y_p$

$y=C_1\cos x+C_2\sin x+C_{3}x\cos x+C_{4}x\sin x+x^2-2x-3$

Hence, the solution is $\mathit{y}\mathbf{=}{\mathit{C}}_{\mathbf{1}}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{+}{\mathit{C}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{+}{\mathit{C}}_{\mathbf{3}}\mathit{x}\mathit{c}\mathit{o}\mathit{s}\mathit{x}\mathbf{+}{\mathit{C}}_{\mathbf{4}}\mathit{x}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{+}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{3}$