Find a particular solution to the differential equation.

$y{}^{\u2033}-{y}^{\prime}-6y=\mathrm{sin}t+3\mathrm{cos}t$

avalg10o
2022-03-25
Answered

Find a particular solution to the differential equation.

$y{}^{\u2033}-{y}^{\prime}-6y=\mathrm{sin}t+3\mathrm{cos}t$

You can still ask an expert for help

Alannah Farmer

Answered 2022-03-26
Author has **11** answers

The given equation is a second order linear non homogeneous differential equation with constant coefficient.

The general solution for$a\left(t\right){y}^{\u2033}+b\left(t\right){y}^{\prime}+c\left(t\right)y=g\left(t\right)$

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 $a\left(t\right){y}^{\u2033}+b\left(t\right){y}^{\prime}+c\left(t\right)y=0$

and the particular solution, is any function that satisfies the non-homogeneous equation.

The complementary solution for the given equation is:

$y{}^{\u2033}-{y}^{\prime}-6y=\mathrm{sin}t+3\mathrm{cos}t$

$y{}^{\u2033}-{y}^{\prime}-6y=0$

$({D}^{2}-D-6)y=0$

${\lambda}^{2}-\lambda -6=0$

$(\lambda -3)(\lambda +2)=0$

$\lambda =3,\lambda =-2$

$y={c}_{1}{e}^{3t}+{c}_{2}{e}^{-2t}$

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:

$({D}^{2}-D-6)y=\mathrm{sin}t+3\mathrm{cos}t$

${y}_{p}=\frac{\mathrm{sin}t+3\mathrm{cos}t}{{D}^{2}-D-6}=\frac{1}{{D}^{2}-D-6}\mathrm{sin}t+3\mathrm{cos}t$

$=\frac{1}{{D}^{2}-D-6}\mathrm{sin}t+\frac{1}{{D}^{2}-D-6}\cdot 3\mathrm{cos}t$

$=\frac{1}{(-1)-D-6}\mathrm{sin}t+\frac{1}{(-1)-D-6}\cdot 3\mathrm{cos}t$

$=\frac{1}{-D-7}\mathrm{sin}t+3\cdot \frac{1}{-D-7}\mathrm{cos}t$

$=-\frac{D-7}{(D+7)(D-7)}\mathrm{sin}t-3\cdot \frac{D-7}{(D-7)(D+7)}\mathrm{cos}t$ {Ratinalizing the denominator}

$=-\frac{D-7}{{D}^{2}-49}\mathrm{sin}t-3\cdot \frac{D-7}{{D}^{2}-49}\mathrm{cos}t$

$=-\frac{D-7}{-1-49}\mathrm{sin}t-3\cdot \frac{D-7}{-1-49}\mathrm{cos}t$

$=\frac{D-7}{50}\mathrm{sin}t+3\cdot \frac{D-7}{50}\mathrm{cos}t$

$=\frac{\mathrm{cos}\left(t\right)}{50}-\frac{7}{50}\mathrm{sin}t+\frac{3(-\mathrm{sin}t)}{50}-\frac{21}{50}\mathrm{cos}t$

$=-\frac{1}{5}\mathrm{sin}\left(t\right)-\frac{2}{5}\mathrm{cos}\left(t\right)$

Now, the solution of the given ODE is:

$y={y}_{p}+{y}_{c}$

$y\left(x\right)={c}_{1}{e}^{3t}+{c}_{2}{e}^{-2t}-\frac{1}{5}\mathrm{sin}\left(t\right)-\frac{2}{5}\mathrm{cos}\left(t\right)$

The general solution for

The general solution of the given differential equation can be written as

and the particular solution, is any function that satisfies the non-homogeneous equation.

The complementary solution for the given equation is:

Two real and different root

The solution is of the form:

So the solution is of the form:

And the particular solution is given as:

Now, the solution of the given ODE is:

Jeffrey Jordon

Answered 2022-03-31
Author has **2262** answers

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${P}_{0}^{{}^{\prime}}\left(t\right)=\alpha {P}_{1}\left(t\right)-\beta {P}_{0}\left(t\right)$

${P}_{1}^{{}^{\prime}}=\beta {P}_{0}\left(t\right)-\alpha {P}_{1}\left(t\right)$

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