Find the general solution of the second order non-homogeneous linear equation: y''-7y'+12y=10sin t+12t+5

The given second order differential equation is $y^{″}-7y^{\prime }+12y=10\sin t+12t+5$
Find the general solution of the differential equation as follows.
The corresponding homogeneous equation is $y^{″}-7y^{\prime }+12y=0$
The auxiliary equation of the corresponding homogeneous equation is
$m^2-7m+12=0$ Obtain the roots of the auxiliary equation as follows.
$m^2-7m+12=0$
$m^2-4m-3m+12=0$
$m(m-4)-3(m-4)=0$
$(m-4)(m-3)=0$
$m=4,m=3$
Therefore, the roots of the auxiliary equation are $m_1=4$ and $m_2=3$
Hence, the complementary solution is $y_c=c_1{e}^{4}t+c_2{e}^{3}t$
Now obtain the particular solution by method of undetermined coefficients as follows.
The choice for the particular solution is $y_p=A\sin (t)+B\cos (t)+Ct+D$
This particular solution $y_p$ satisfies the given differential equation.
Therefore,
$y_p^{″}-7y_p^{\prime }+12y_p=10\sin t+12t+5$
$-A\sin (t)-B\cos (t)-7(A\cos (t)-B\sin (t)+C)$
$+12(A\sin (t)+B\cos (t)+Ct+D)=10\sin t+12t+5$
$-A\sin (t)-B\cos (t)-7A\cos (t)+7B\sin (t)-7C+12A\sin (t)$
$+12B\cos (t)+12Ct+12D=10\sin t+12t+5$
$(7B+11A)\sin (t)+(11B-7A)\cos (t)+12Ct+12D-7C=10\sin t+12t+5$
Equate the coefficients of like terms on both sides and obtain,
$7B+11A=10$ (1)
$11B-7A=0$ (2)
$12C=12$ (3)
$12D-7C=5$ (4) Solve the first two equations as shown below
$(1)\cdot 11\Rightarrow 77B+121A=110$
$(2)\cdot -7\Rightarrow -77B+49A=0$
$170A=110$
Then, $A=\frac{110}{170}=\frac{11}{17}$
Substitute $A=\frac{11}{17}$ in the equation $7B+11A=-10$ and solve for B as follows.
$7B+11A=10$
$7B+11(\frac{11}{17})=10$
$7B+\frac{121}{17}=10$
$7B=\frac{170-121}{17}$
$7B=\frac{49}{17}$
$B=\frac{7}{17}$
Now from third equation $12C=12$, we have $C=1$.
Then,
$12D-7C=5$
$12D-7=5$
$12D=12$
$D=1$
Hence the particular solution is,
$y_p=A\sin (t)+B\cos (t)+Ct+D$

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