The given second order differential equation is \(y''-7y'+12y=10\sin t+12t+5\)

Find the general solution of the differential equation as follows.

The corresponding homogeneous equation is \(y''-7y'+12y=0\)

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 \(y_c=c_1e^4t+c_2e^3t\)

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,

\(y_p''-7y_p'+12y_p=10\sin t+12t+5\)

\(-A\sin(t)-B\cos(t)-7(A\cos(t)-B\sin(t)+C)\)

\(+12(A\sin(t)+B\cos(t)+Ct+D)=10\sin t+12t+5\)

\(-A\sin(t)-B\cos(t)-7A\cos(t)+7B\sin(t)-7C+12A\sin(t)\)

\(+12B\cos(t)+12Ct+12D=10\sin t+12t+5\)

\((7B+11A)\sin(t)+(11B-7A)\cos(t)+12Ct+12D-7C=10\sin t+12t+5\)

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

\((1)\cdot11\Rightarrow77B+121A=110\)

\((2)\cdot-7\Rightarrow-77B+ 49A=0\)

\(170A=110\)

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,

\(y_p=A\sin(t)+B\cos(t)+Ct+D\)

\(=\frac{11}{17}\sin(t)+\frac{7}{17}\cos(t)+t+1\)

Therefore the general solution of the differential equation

\(y''-7y'+12y=10\sin t+12t+5\) is,

\(y=y_c+y_p\)

\(=c_1e^{4t}+c_2e^{3t}+\frac{11}{17}\sin(t)+\frac{7}{17}\cos(t)+t+1\)

Thus, the general solution of the differential equation is,

\(y=c_1e^{4t}+c_2e^{3t}+\frac{11}{17}\sin(t)+\frac{7}{17}\cos(t)+t+1\)