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

Find the general solution of the second order non-homogeneous linear equation:
${y}^{″}-7{y}^{\prime }+12y=10\mathrm{sin}t+12t+5$
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Willie

The given second order differential equation is ${y}^{″}-7{y}^{\prime }+12y=10\mathrm{sin}t+12t+5$
Find the general solution of the differential equation as follows.
The corresponding homogeneous equation is ${y}^{″}-7{y}^{\prime }+12y=0$
The auxiliary equation of the corresponding homogeneous equation is
${m}^{2}-7m+12=0$ Obtain the roots of the auxiliary equation as follows.
${m}^{2}-7m+12=0$
${m}^{2}-4m-3m+12=0$
$m\left(m-4\right)-3\left(m-4\right)=0$
$\left(m-4\right)\left(m-3\right)=0$

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 ${y}_{p}=A\mathrm{sin}\left(t\right)+B\mathrm{cos}\left(t\right)+Ct+D$
This particular solution ${y}_{p}$ satisfies the given differential equation.
Therefore,
${y}_{p}^{″}-7{y}_{p}^{\prime }+12{y}_{p}=10\mathrm{sin}t+12t+5$
$-A\mathrm{sin}\left(t\right)-B\mathrm{cos}\left(t\right)-7\left(A\mathrm{cos}\left(t\right)-B\mathrm{sin}\left(t\right)+C\right)$
$+12\left(A\mathrm{sin}\left(t\right)+B\mathrm{cos}\left(t\right)+Ct+D\right)=10\mathrm{sin}t+12t+5$
$-A\mathrm{sin}\left(t\right)-B\mathrm{cos}\left(t\right)-7A\mathrm{cos}\left(t\right)+7B\mathrm{sin}\left(t\right)-7C+12A\mathrm{sin}\left(t\right)$
$+12B\mathrm{cos}\left(t\right)+12Ct+12D=10\mathrm{sin}t+12t+5$
$\left(7B+11A\right)\mathrm{sin}\left(t\right)+\left(11B-7A\right)\mathrm{cos}\left(t\right)+12Ct+12D-7C=10\mathrm{sin}t+12t+5$
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
$\left(1\right)\cdot 11⇒77B+121A=110$
$\left(2\right)\cdot -7⇒-77B+49A=0$
$170A=110$
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.
$7B+11A=10$
$7B+11\left(\frac{11}{17}\right)=10$
$7B+\frac{121}{17}=10$
$7B=\frac{170-121}{17}$
$7B=\frac{49}{17}$
$B=\frac{7}{17}$
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,
${y}_{p}=A\mathrm{sin}\left(t\right)+B\mathrm{cos}\left(t\right)+Ct+D$