Toronto Math Forum
MAT2442018F => MAT244Lectures & Home Assignments => Topic started by: chloe on October 14, 2018, 12:28:15 AM

y''+y'=3sin2t+tcos2t

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
$$Y(t)=(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}.$$
$$Y'(t)=(A_0B_1)\sin{2t}+(2A_1+B_0)\cos{2t}2B_0 t\sin{2t}+2A_1 t\cos{2t}$$
$$Y''(t)=2(A_0 2 B_1)\cos{2t}+2A_0\cos{2t}4A_0 t\sin{2t}2(2A_1+B_0)\sin{2t}2B_0\sin{2t}4B_0t\cos{2t}$$
Plugging $Y$, and $Y''$ into $(1)$, we have
$$[(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}]''+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$2(A_0 2 B_1)\cos{2t}+2A_0\cos{2t}4A_0 t\sin{2t}2(2A_1+B_0)\sin{2t}2B_0\sin{2t}4B_0t\cos{2t}+(A_0 t+A_1)\sin{2t}+(B_0 t+B_1)\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$(4A_03B_1)\cos{2t}+(3A_0)t\sin{2t}+(3A_14B_0)\sin{2t}+(3B_0)t\cos{2t}=3\sin{2t}+t\cos{2t}$$
$$\left\{\begin{aligned} &4A_03B_1=0\\ &3A_0=0\\& 3A_14B_0=3\\&3B_0=1\end{aligned}\right. \label{eq4} $$
$$\Rightarrow A_0=0, A_1=\frac{5}{9}, B_0=\frac{1}{3}, B_1=0$$
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

typing that out was brutal, so I hope this helps lol
If you think this was brutal, you are a wimp :D
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
$$
Y'= (2B_0t+2B_1+\underline{A_0})\cos(2t)+(2A_0t2A_1+\underline{B_0})\sin(2t)
$$
where underlined terms are obtained by the differentiation of the polynomial factors and other terms by the differentiation of the trigonometric factors.
Further
$$
Y''= (4A_0t4A_1 + \underline{4B_0} + \underbracket{0} )\cos(2t)+(4B_0t4B_1 \underline{4A_0}+\underbracket{0} )\sin(2t)
$$
where underlined terms are obtained by the 1time differentiation of the polynomial factors and 1time differentiation of the trigonometric factors (you need to double it), and other terms by the 2times differentiation of the trigonometric factors. I also included $ \underbracket{0}$ obtained by 2times differentiation of the polynomial factors