Find the general solution of y^((4)) + 2y''' + 5y'' + 8y' +…

Find the general solution of $y^{(4)} + 2y''' + 5y'' + 8y' + 4y = \cos x + 40e^{x}$.

Step 1: Characteristic equation

Substituting $y=e^{rx}$ into the homogeneous equation gives the quartic

\[ r^4+2r^3+5r^2+8r+4=0. \]

Step 2: Finding a rational root

Testing $r=-1$: $1-2+5-8+4=0.$ $\checkmark$

Dividing out $(r+1)$ gives $r^4+2r^3+5r^2+8r+4=(r+1)(r^3+r^2+4r+4)$.

Step 3: Factoring the cubic by grouping

\[ r^3+r^2+4r+4 = r^2(r+1)+4(r+1) = (r+1)(r^2+4). \] So altogether

\[ r^4+2r^3+5r^2+8r+4 = (r+1)^2(r^2+4). \]

Check by expansion: $(r+1)^2(r^2+4)=(r^2+2r+1)(r^2+4)=r^4+4r^2+2r^3+8r+r^2+4=r^4+2r^3+5r^2+8r+4. $\checkmark$$

Step 4: Roots and homogeneous solution

The roots are $r=-1$ (double) and $r=\pm2i$, so

\[ Y_h = (C_1+C_2x)e^{-x} + C_3\cos2x + C_4\sin2x. \]

Step 5: Superposition

By the superposition principle, split $y^{*}=y_1^{*}+y_2^{*}$, where $y_1^{*}$ handles $\cos x$ and $y_2^{*}$ handles $40e^x$.

Step 6: Part 1: $f_1(x)=\cos x$

Here $\alpha=0$, $\beta=1$, complex exponent $i$. Comparing against the roots $\{-1(\text{double}),\pm2i\}$: $i$ is not a root, so $s=0$ and the trial is

\[ y_1^{*}=A\cos x+B\sin x. \]

Its derivatives are

\[ y_1^{*\prime}=-A\sin x+B\cos x,\quad y_1^{*\prime\prime}=-A\cos x-B\sin x,\quad y_1^{*\prime\prime\prime}=A\sin x-B\cos x,\quad y_1^{*(4)}=A\cos x+B\sin x. \]

Substituting into the full equation and collecting the $\cos x$ and $\sin x$ coefficients: $\cos x:\ A-2B-5A+8B+4A = 6B, \qquad \sin x:\ B+2A-5B-8A+4B = -6A.$

Matching against $1\cdot\cos x+0\cdot\sin x$: $6B=1\Rightarrow B=\dfrac16$, and $-6A=0\Rightarrow A=0$. So

\[ y_1^{*}=\frac{\sin x}{6}. \]

Step 7: Part 2: $f_2(x)=40e^{x}$

Here $\alpha=1$, which is not among the roots $\{-1,\pm2i\}$, so $s=0$ and the trial is $y_2^{*}=Ce^{x}$. Since every derivative of $e^x$ equals $e^x$, substituting into the equation gives

\[ C e^{x}(1+2+5+8+4) = 40e^{x} \implies 20C=40 \implies C=2. \] So $y_2^{*}=2e^{x}$.

Step 8: General solution

The solution is recorded in the result below.