Question

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

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

2021-02-03

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