Toronto Math Forum

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*}

Find general and singular solutions to:
                                                                                $2y-4xy'-\ln(y')=0$

1. We transform the equation into the form:
                                                                                $y=x\cdot 2y'+\frac{1}{2}\ln(y')$

We plug $p=y'$ and differentiate the equation, we get:
                                                                       
                                                                                $pdx = 2pdx+2x\cdot dp+\frac{1}{2}\frac{1}{p}\cdot dp$

Then:
                                                                                $p\frac{dx}{dp}=2p\frac{dx}{dp}+2x+\frac{1}{2p}$
                                                       
                                                                                $-px'-2x=\frac{1}{2p}$
                                                       
                                                                                $x'+\frac{2}{p}x=-\frac{1}{2p^2}$

Then $p(p) = \frac{2}{p}$, $u(p)=e^{2\int \frac{1}{p}dp} = e^{2lnp}=p^2$,
Multiply both sides with integrating factor $u(p) = p^2$ and we get:
                                                       
                                                                                $p^2x'+2px=-\frac{1}{2}$
 
                                                                                $(p^2x)'=-\frac{1}{2}$

                                                                                $p^2x=-\int \frac{1}{2}dp$

                                                                                $p^2x=-\frac{1}{2}p+C$

                                                                                $x=-\frac{1}{2p}+\frac{C}{p^2}$

Then the solution in the parametric form is: $\begin{cases}x=-\frac{1}{2p} + \frac{C}{p^2}\\ y=x\cdot 2p + \frac{1}{2}\ln p\end{cases}$


2. To find the singular solution, we solve the equation $\varphi(p) -p=0 \implies 2p-p=0 \implies p=0$

It follows from this that $y=C$, we can make direct substitution to make sure that the constant $C$ is equal to 0.

Therefore, the differential equation has the singular solution $y=0$.                                       

Question: $ty^{'}+2y=sin(t)$,   $t>0$

standard equation form: $y^{'}+\frac{2}{t}y=\frac{sin(t)}{t}$

$p(t) = \frac{2}{t}$,  $g(t) = \frac{sin(t)}{t}$

$u = e^{\int p(t)}dt = e^{2\int \frac{1}{t}dt} = t^2$, then we multiply both sides with $u$, and we get:

$t^2y^{'} + 2ty = tsin(t)$

$(t^2y)^{'}=tsin(t)$

$d(t^2y) = tsin(t)dt$

$t^2y = \int tsin(t) dt$

(integrating by parts, $u=t \implies du =dt$ and $dv=sin(t)$ and $v=-cos(t)$

$\int tsin(t)dt = uv - \int vdu$

                   $=-tcos(t)-\int (-cos(t))dt$
                   
                   $=-tcos(t) +\int cos(t)dt$

                   $=-tcos(t) +sin(t) + C$)

Therefore, $t^2y=-tcos(t)+sin(t) + C$

$y=\frac{-tcos(t)+sin(t)+C}{t^2}$.

Since $t>0$, and when $t \rightarrow \infty, y \rightarrow 0$

tut0401   $y^{'} + \frac{2}{3}y=1-\frac{1}{2}t$, $y(0) = y_0$
   
$p(t) = \frac{2}{3}$, $g(t) = 1-\frac{1}{2}t$, then $u = e^{\int \frac{2}{3}dt}$,

multiply both sides with $u$, then we get:

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

$(e^{\frac{2}{3}t}y)^{'}=e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t}$

$d(e^{\frac{2}{3}t}y) = (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$

$e^{\frac{2}{3}t}y=\int (e^{\frac{2}{3}t}-\frac{1}{2}te^{\frac{2}{3}t})dt$

$e^{\frac{2}{3}t}y=\frac{3}{2}e^{\frac{2}{3}t}-\frac{3}{4}te^{\frac{2}{3}t}+\frac{9}{8}e^{\frac{2}{3}t}+C$    (integrate by parts)

$y= \frac{3}{2}-\frac{3}{4}t+\frac{9}{8}+Ce^{-\frac{2}{3}t}$

$y=\frac{21}{8}-\frac{3}{4}t+Ce^{-\frac{2}{3}t}$. Consider $y(0) = y_0 \implies y_0=\frac{21}{8}-0+C \implies C=y_0-\frac{21}{8}$

Therefore, $y=\frac{21}{8}-\frac{3}{4}t+(y_0-\frac{21}{8})e^{-\frac{2}{3}t}$

If $y(t)$ touches, but does not cross the $t$-axis at some point $t_0$ s.t. $y(t_0) = 0$ and $y^{'}(t_0) = 0$

Thus, $y^{'} +\frac{2}{3}y = 1- \frac{1}{2}t$ when $t=t_0$

$0+0 = 1-\frac{1}{2}t_0$

$\frac{1}{2}t_0 = 1 \implies t_0 = 2$

substitute $t_0 = 2$, then $y(2) = 0 \implies 0 = \frac{21}{8} - \frac{3}{4}\times 2 + (y_0-\frac{21}{8})e^{-\frac{4}{3}}$

Therefore, we get $y_0 = \frac{21}{8}e^{-\frac{4}{3}}-\frac{9}{8}$