Author Topic: Q4 TUt 0401  (Read 1040 times)

Victor Ivrii

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Q4 TUt 0401
« on: October 26, 2018, 05:43:09 PM »
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given inhomogeneous equation.
\begin{gather*}
t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3,\qquad t > 0;\\
y_1(t) = t,\quad y_2(t) = te^t.
\end{gather*}

Monika Dydynski

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Re: Q4 TUt 0401
« Reply #1 on: October 26, 2018, 06:22:16 PM »
Verify that $y_1(t)$ and $y_2(t)$ satisfy the corresponding homogeneous equation, then find a particular solution of the given nonhomogeneous equation,

$$\begin{gather*} t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3,\qquad t > 0;\\ y_1(t) = t,\quad y_2(t) = te^t. \end{gather*}$$

1. Verify that $y_1(t)=t$ and $y_2(t)=te^{t}$ satisfy the corresponding homogeneous equation,

$$t^2y'' - t(t + 2)y' + (t + 2)y =0.\tag{2}$$

$$\cases{y_1(t)=t\\y'_1(t)=1\\y''_1(t)=0}$$

Plugging $y_1$, $y'_1$, and $y''_1$ into $(2)$,

$$t^2(0)- t(t + 2)(1) + (t + 2)(t) =0 \Rightarrow -t^{2}-2t+t^{2}+2t=0$$

$y_1$ satisfies $(2)$ $\Rightarrow$ $y_1$ is a solution to the corresponding homogeneous equation.


Similarily,

$$\cases{y_2(t)=te^{2}\\y'_2(t)=te^{t}+e^{t}\\y''_2(t)=te^{t}+2e^{t}}$$

Plugging $y_2$, $y'_2$, and $y''_2$ into $(2)$,

$$t^2(te^t+2e^t) - t(t + 2)(te^t+e^t) + (t + 2)(te^t) =0 \Rightarrow t^3e^t-t^3e^t+3t^2e^t-3t^2e^t+2te^t-2te^t=0$$

$y_2$ satisfies $(2)$ $\Rightarrow$ $y_2$ is a solution to the corresponding homogeneous equation.


2. Find a particular solution of the given nonhomogeneous equation,

$$t^2y'' - t(t + 2)y' + (t + 2)y = 2t^3, t>0.\tag{1}$$

Dividing $(1)$ by $t^{2}$, we have

$$y''-\frac{t+2}{t}y'+\frac{t+2}{t^{2}}=2t$$

Note that $p(t)=-1-\frac{2}{t}$, $q(t)=\frac{1}{t}+\frac{2}{t^2}$, and $g(t)=2t$ are continuous on $(0, +\infty)$

Computing the Wronskian,

$$W(t, te^{t})(t)=\begin{array}{|c c|}t& te^{t} \\ 1 & te^t+e^t\end{array}=t^2e^t\ne0,$$

we verify that $y_1(t)=t$ and $y_2(t)=te^{t}$ form a fundamental set of solutions $\forall t>0$

Calculating the parameters $u_1$ and $u_2$, we get

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\frac{te^{t}2t}{t^2e^t}}dt=-\int{2}dt=-2t+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{\frac{t2t}{t^2e^t}}dt=2\int{e^{-t}}dt=-2e^{-t}+c_2$$

The particular solution is

$$y_p(t)=t(-2t+c_1)+te^{t}(-\frac{2}{e^t}+c_2)=-2t^{2}-2t+tc_1+te^tc_2$$

$$y_p(t)=-2t^{2}.$$
« Last Edit: October 26, 2018, 06:25:44 PM by Monika Dydynski »

Chengyin Ye

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Re: Q4 TUt 0401
« Reply #2 on: October 26, 2018, 06:22:39 PM »
Here is my solution.