Question

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

Answer 1

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