Author Topic: Q4 TUT0101  (Read 968 times)

Victor Ivrii

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Q4 TUT0101
« on: October 26, 2018, 05:40:42 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''-2y=3t^2-1,\qquad t>0;\\
 y_1(t)=t^2,\quad  y_2(t)=t^{-1}.
\end{gather*}

Monika Dydynski

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Re: Q4 TUT0101
« Reply #1 on: October 26, 2018, 05:43:58 PM »
Verify that the given functions $y_1$ and $y_2$ satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.

$$\begin{gather*} t^2y''-2y=3t^2-1,\qquad t>0\tag{1};\\  y_1(t)=t^2,\quad  y_2(t)=t^{-1}. \end{gather*}$$

1. Verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ satisfy the corresponding homogeneous equation,

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

$$\cases{y_1(t)=t^{2}\\y'_1(t)=2t\\y''_1(t)=2}$$

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

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

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


Similarily,

$$\cases{y_2(t)=t^{-1}\\y'_2(t)=-t^{-2}\\y''_2(t)=2t^{-3}}$$

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

$$t^{2}\left(\frac{2}{t^{3}}\right)-2\left(\frac{1}{t}\right)=0 \Rightarrow \frac{2}{t}-\frac{2}{t}=0$$

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


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

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

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

$$y''-\frac{2}{t^{2}}y=3-\frac{1}{t^{2}}$$

Note that $p(t)=0$, $q(t)=-\frac{2}{t^{2}}$, and $g(t)=3-\frac{1}{t^{2}}$ are continuous on $(0, +\infty)$

Computing the Wronskian,

$$W(t^{2}, t^{-1})(t)=\begin{array}{|c c|}t^{2}& t^{-1} \\ 2t & -t^{-2}\end{array}=-1-2=-3\ne0,$$

we verify that $y_1(t)=t^{2}$ and $y_2(t)=t^{-1}$ form a fundamental set of solutions $\forall t>0$

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

$$u_1=-\int{\frac{y_2(t)g(t)}{W(y_1,y_2)(t)}}dt=-\int{\frac{t^{-1}(3-t^{-2})}{-3}}dt=\frac{1}{3}\int{\frac{3}{t}-\frac{1}{t^{3}}}dt=\ln{t}+\frac{1}{6t^{2}}+c_1$$

$$u_2=\int{\frac{y_1(t)g(t)}{W(y_1,y_2)(t)}}dt=\int{\frac{t^{2}(3-t^{-2})}{-3}}dt=-\frac{1}{3}\int{3t^{2}-1}dt=-\frac{t^{3}}{3}+\frac{t}{3}+c_2$$

The particular solution is

$$y_p(t)=t^{2}\left(\ln{t}+\frac{1}{6t^{2}}+c_1\right)+\frac{1}{t}\left(-\frac{t^{3}}{3}+\frac{t}{3}+c_2\right)=t^{2}\ln{t}+\frac{1}{6}+c_1t^{2}-\frac{t^{2}}{3}+\frac{1}{3}+\frac{c_2}{t}=t^{2}\ln{t}+\frac{1}{2}+c_1t^{2}+\frac{c_2}{t}$$

$$y_p(t)=t^{2}\ln{t}+\frac{1}{2}.$$
« Last Edit: October 26, 2018, 05:46:37 PM by Monika Dydynski »