### Author Topic: non-homogeneous question  (Read 1645 times)

#### chloe

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##### non-homogeneous question
« on: October 14, 2018, 12:28:15 AM »
y''+y'=3sin2t+tcos2t

#### Monika Dydynski

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##### Re: non-homogeneous question
« Reply #1 on: October 14, 2018, 02:43:05 AM »
I'm assuming this is question 10 from 3.5, in which case you've copied the question wrong. The given DE should read:
$$y''+y=3\sin{2t}+t\cos{2t}\tag{1}.$$
The corresponding homogeneous equation is
$$y''+y=0$$
with characteristic equation
$$r^{2}+1=0$$
and roots
$$r_{1,2}=\pm i$$.
The solution to the corresponding homogenous equation is
$$y_h=c_1 \cos{t}+c_2 \sin{t}.$$
Particular Solution (using method of undetermined coefficients):

Splitting up the right side of $(1)$, we get two equations
$$y''+y=3\sin{2t},$$
$$y''+y=t\cos{2t}.$$
Since $\sin{t}$ and $\cos{t}$ are not solutions to the corresponding homogenous equation, we set our particular solution to be of the form
$$Y(t)=(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}.$$
$$Y'(t)=(A_0-B_1)\sin{2t}+(2A_1+B_0)\cos{2t}-2B_0 t\sin{2t}+2A_1 t\cos{2t}$$
$$Y''(t)=2(A_0 -2 B_1)\cos{2t}+2A_0\cos{2t}-4A_0 t\sin{2t}-2(2A_1+B_0)\sin{2t}-2B_0\sin{2t}-4B_0t\cos{2t}$$
Plugging $Y$, and $Y''$ into $(1)$, we have
$$[(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}]''+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$2(A_0 -2 B_1)\cos{2t}+2A_0\cos{2t}-4A_0 t\sin{2t}-2(2A_1+B_0)\sin{2t}-2B_0\sin{2t}-4B_0t\cos{2t}+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$(4A_0-3B_1)\cos{2t}+(-3A_0)t\sin{2t}+(-3A_1-4B_0)\sin{2t}+(-3B_0)t\cos{2t}=3\sin{2t}+t\cos{2t}$$

\left\{\begin{aligned} &4A_0-3B_1=0\\ &-3A_0=0\\& -3A_1-4B_0=3\\&-3B_0=1\end{aligned}\right. \label{eq4}

$$\Rightarrow A_0=0, A_1=-\frac{5}{9}, B_0=-\frac{1}{3}, B_1=0$$

Thus, the particular solution of the nonhomogenous equation is

$$Y(t)=-\frac{5}{9} \sin{2t}-\frac{1}{3} t\cos{2t}.$$

The solution to $(1)$ is

$$y(t)=y_h(t)+Y(t)=c_1\cos{t}+c_2\sin{t}-\frac{5}{9} \sin{2t}-\frac{1}{3} t\cos{2t}.$$

typing that out was brutal, so I hope this helps lol

« Last Edit: October 14, 2018, 06:53:10 AM by Victor Ivrii »

#### Victor Ivrii

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##### Re: non-homogeneous question
« Reply #2 on: October 14, 2018, 06:33:32 AM »
Quote
typing that out was brutal, so I hope this helps lol
If you think this was brutal, you are a wimp

Actually, there is a shorter way to differentiate expression like this:
$$Y = (A_0t+A_1)\cos(2t)+(B_0t+B_1)\sin(2t).$$
Then
$$Y'= (2B_0t+2B_1+\underline{A_0})\cos(2t)+(-2A_0t-2A_1+\underline{B_0})\sin(2t)$$
where underlined terms are obtained by the differentiation of the polynomial factors and other terms by the differentiation of the trigonometric factors.
Further
$$Y''= (-4A_0t-4A_1 + \underline{4B_0} + \underbracket{0} )\cos(2t)+(-4B_0t-4B_1 \underline{-4A_0}+\underbracket{0} )\sin(2t)$$
where underlined terms are obtained by the 1-time differentiation of the polynomial factors and 1-time differentiation of the trigonometric factors (you need to double it), and other terms by the 2-times differentiation of the trigonometric factors. I also included $\underbracket{0}$ obtained by 2-times differentiation of the polynomial factors
« Last Edit: October 14, 2018, 07:02:49 AM by Victor Ivrii »