Author Topic: Q4 TUT 0501  (Read 1135 times)

Victor Ivrii

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Q4 TUT 0501
« on: October 26, 2018, 05:43:41 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*}
x^2y'' + xy' + (x^2 - 0.25)y = 3x^{3/2} \sin (x),\qquad x > 0;\\
y_1(x) = x^{-1/2} \sin (x),\quad y_2(x) = x^{-1/2} \cos (x).
\end{gather*}

Jiabei Bi

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Re: Q4 TUT 0501
« Reply #1 on: October 26, 2018, 08:59:47 PM »
The file is too large to upload. So I upload two files.

Wei Cui

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Re: Q4 TUT 0501
« Reply #2 on: October 27, 2018, 11:58:12 AM »
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*}
x^2y'' + xy' + (x^2 - 0.25)y = 3x^{3/2} \sin (x),\qquad x > 0;\\
y_1(x) = x^{-1/2} \sin (x),\quad y_2(x) = x^{-1/2} \cos (x).
\end{gather*}


Hence,
\begin{gather*}
\begin{cases}
y_1(x) = x^{-\frac{1}{2}}sin(x)\\
y_1'(x) = -\frac{1}{2}x^{-\frac{3}{2}}sin(x) + x^{-\frac{1}{2}}cos(x)\\
y_1''(x) = \frac{3}{4}x^{-\frac{5}{2}}sin(x)-x^{-\frac{3}{2}}cos(x)-x^{-\frac{1}{2}}sin(x)
\end{cases}
\end{gather*}



\begin{gather*}
\begin{cases}
y_2(x) = x^{-\frac{1}{2}}cos(x)\\
y_2'(x) = -\frac{1}{2}x^{-\frac{3}{2}}cos(x) - x^{-\frac{1}{2}}sin(x)\\
y_2''(x) = \frac{3}{4}x^{-\frac{5}{2}}cos(x) + x^{-\frac{3}{2}}sin(x)-x^{-\frac{1}{2}}cos(x)
\end{cases}
\end{gather*}

Substitute back into the homogeneous equation:
\begin{gather*}
x^2y'' + xy' + (x^2 - 0.25)y = 0
\end{gather*}

Verified that $y_1(x)$ and $y_2(x)$ both satisfy the corresponding homogeneous equation.
And the complementary solution $y_c(x) = c_1x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x)$
Now divide both sides of the original equation by $x^2$:
\begin{gather*}
y'' + \frac{1}{x}y' + \frac{x^2-0.25}{x^2}y = 3x^{-\frac{1}{2}}sin(x)
\end{gather*}

Then
\begin{gather*}
p(t) = \frac{1}{x}, q(t) = \frac{x^2 - 0.25}{x^2}, g(t) = 3x^{-\frac{1}{2}}sin(x)\\
\ \\
W[y_1,y_2](x) =
\begin{vmatrix}
y_1(x) & y_2(x)\\
y'_1(x) & y'_2(x)
\end{vmatrix}
= -x^{-1} \neq 0
\end{gather*}


Since the particular solution has the form:
\begin{gather*}
Y_p(x) = u_1(x)y_1(x) + u_2(x)y_2(x)
\end{gather*}

and
\begin{align*}
u_1(x) &= - \int \frac{y_2(x) g(x)}{W[y_1,y_2](x)}dx\\
&= -\int \frac{x^{-\frac{1}{2}}cos(x)\cdot 3x^{-\frac{1}{2}}sin(x)}{-x^{-1}} dx\\
&= \frac{3}{2}\int 2sin(x)cos(x) dx\\
&= \frac{3}{2}\int sin(2x)dx\\
&= -\frac{3}{4}cos(2x)
\end{align*}



\begin{align*}
u_2(x) &= \int \frac{y_1(x) g(x)}{W[y_1,y_2](x)}dx\\
&= \int \frac{x^{-\frac{1}{2}}sin(x)\cdot 3x^{-\frac{1}{2}}sin(x)}{-x^{-1}} dx\\
&= -3\int sin^2(x) dx\\
&= -3\int \frac{1-cos(2x)}{2}dx\\
&= -\frac{3}{2}x + \frac{3}{4}sin(2x)
\end{align*}


Therefore,

\begin{align*}
Y_p(x) &= -\frac{3}{4}cos(2x)\cdot x^{-\frac{1}{2}} sin(x) + (-\frac{3}{2}x + \frac{3}{4}sin(2x))\cdot x^{-\frac{1}{2}}cos(x)\\
&= \frac{3}{4}x^{-\frac{1}{2}}(sin(2x)cos(x)-cos(2x)sin(x))- \frac{3}{2}x^{-\frac{1}{2}}cos(x)\\
&= \frac{3}{4}x^{-\frac{1}{2}}(2sin(x)cos^2(x)-(2cos^2(x)- cos(x))sin(x))-\frac{3}{2}x^{\frac{1}{2}}cos(x)\\
&= \frac{3}{4}x^{-\frac{1}{2}}sin(x) - \frac{3}{2}x^{\frac{1}{2}}cos(x)
\end{align*}


Hence, the general solution:
\begin{align*}
y(x) &= y_c(x) + Y_p(x)\\
&= c_1x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x) + \frac{3}{4}x^{-\frac{1}{2}}sin(x) - \frac{3}{2}x^{\frac{1}{2}}cos(x)\\
&= (c_1 + \frac{3}{4})x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x) - \frac{3}{2}x^{\frac{1}{2}}cos(x)\\
&= c_3x^{-\frac{1}{2}}sin(x) + c_2x^{-\frac{1}{2}}cos(x)-\frac{3}{2}x^{\frac{1}{2}}cos(x)
\end{align*}

where $c_3 = c_1 + \frac{3}{4}$
Therefore, the particular solution of the given non homogenous equation is
\begin{gather*}
Y_p(x) = -\frac{3}{2}x^{\frac{1}{2}}cos(x)
\end{gather*}
« Last Edit: October 27, 2018, 12:08:02 PM by Wei Cui »