# Find a particular solution to the differential equation.

Find a particular solution to the differential equation.
$y{}^{″}-{y}^{\prime }-6y=\mathrm{sin}t+3\mathrm{cos}t$
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Alannah Farmer
The given equation is a second order linear non homogeneous differential equation with constant coefficient.
The general solution for $a\left(t\right){y}^{″}+b\left(t\right){y}^{\prime }+c\left(t\right)y=g\left(t\right)$
The general solution of the given differential equation can be written as
$y={y}_{h}+{y}_{p}$
${y}_{h}$ is the solution to the homogeneous ODE $a\left(t\right){y}^{″}+b\left(t\right){y}^{\prime }+c\left(t\right)y=0$
and the particular solution, is any function that satisfies the non-homogeneous equation.
The complementary solution for the given equation is:
$y{}^{″}-{y}^{\prime }-6y=\mathrm{sin}t+3\mathrm{cos}t$
$y{}^{″}-{y}^{\prime }-6y=0$
$\left({D}^{2}-D-6\right)y=0$
${\lambda }^{2}-\lambda -6=0$
$\left(\lambda -3\right)\left(\lambda +2\right)=0$
$\lambda =3,\lambda =-2$
$y={c}_{1}{e}^{3t}+{c}_{2}{e}^{-2t}$
Two real and different root ${\lambda }_{1}$ and ${\lambda }_{2}$
The solution is of the form: $y={c}_{1}{e}^{{\lambda }_{1}\cdot t}+{c}_{2}{e}^{{\lambda }_{2}\cdot t}$
So the solution is of the form:
${y}_{c}={c}_{1}{e}^{3t}+{c}_{2}{e}^{-2t}$
And the particular solution is given as:
$\left({D}^{2}-D-6\right)y=\mathrm{sin}t+3\mathrm{cos}t$
${y}_{p}=\frac{\mathrm{sin}t+3\mathrm{cos}t}{{D}^{2}-D-6}=\frac{1}{{D}^{2}-D-6}\mathrm{sin}t+3\mathrm{cos}t$
$=\frac{1}{{D}^{2}-D-6}\mathrm{sin}t+\frac{1}{{D}^{2}-D-6}\cdot 3\mathrm{cos}t$
$=\frac{1}{\left(-1\right)-D-6}\mathrm{sin}t+\frac{1}{\left(-1\right)-D-6}\cdot 3\mathrm{cos}t$
$=\frac{1}{-D-7}\mathrm{sin}t+3\cdot \frac{1}{-D-7}\mathrm{cos}t$
$=-\frac{D-7}{\left(D+7\right)\left(D-7\right)}\mathrm{sin}t-3\cdot \frac{D-7}{\left(D-7\right)\left(D+7\right)}\mathrm{cos}t$ {Ratinalizing the denominator}
$=-\frac{D-7}{{D}^{2}-49}\mathrm{sin}t-3\cdot \frac{D-7}{{D}^{2}-49}\mathrm{cos}t$
$=-\frac{D-7}{-1-49}\mathrm{sin}t-3\cdot \frac{D-7}{-1-49}\mathrm{cos}t$
$=\frac{D-7}{50}\mathrm{sin}t+3\cdot \frac{D-7}{50}\mathrm{cos}t$
$=\frac{\mathrm{cos}\left(t\right)}{50}-\frac{7}{50}\mathrm{sin}t+\frac{3\left(-\mathrm{sin}t\right)}{50}-\frac{21}{50}\mathrm{cos}t$
$=-\frac{1}{5}\mathrm{sin}\left(t\right)-\frac{2}{5}\mathrm{cos}\left(t\right)$
Now, the solution of the given ODE is:
$y={y}_{p}+{y}_{c}$
$y\left(x\right)={c}_{1}{e}^{3t}+{c}_{2}{e}^{-2t}-\frac{1}{5}\mathrm{sin}\left(t\right)-\frac{2}{5}\mathrm{cos}\left(t\right)$
Jeffrey Jordon