Toronto Math Forum

5. Verify that the given functions y1 and y2 satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation.\\

\begin{eqnarray*}
&& (1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1 \\
&& y_{1}(t)=e^{t}, y_{2}(t)=t.
\end{eqnarray*}


$$(1-t)y''+ty'-y=2(t-1)^{2}e^{-t}, 0<t<1;y_{1}(t)=e^{t}, y_{2}(t)=t$$ 
Hence,

$\left\{
\begin{array}{l}
y_{1}(t)=e^{t} \\
y_{1}'(t)=e^{t} \\
y_{1}''(t)=e^{t}
\end{array}
\right.$
and
$\left\{
\begin{array}{l}
y_{2}(t)=t\\
y_{2}'(t)=1\\
y_{2}''(t)=0
\end{array}
\right.$

Substitute back into the homogeneous equation:
$$(1-t)y''+ty'-y=0$$
\ \\
Verified that $y_{1}(t)$ and $y_{2}(t)$ both satisfy the corresponding homogeneous equation.\\
And the complementary solution $y_{c}(t)=c_{1}e^{t}+c_{2}$\\
Now divide both sides of the original equation by $1-t$:\\
$$y''+\dfrac{t}{1-t}-\dfrac{1}{1-t}=-2(t-1)e^{-t}$$
\ \\
Then\\
$$p(t)=\dfrac{t}{1-t},q(t)=-\dfrac{1}{1-t},g(t)=-2(t-1)e^{-t}$$
$$W[y_{1},y_{2}](t)=
\left|
\begin{array}{cc}
y_{1}(t) & y_{2}(t) \\
y_{1}'(t) & y_{2}'(t)
\end{array}
\right|
=(1-t)e^{t}$$
Since the particular solution has the form:\\
$$Y(t)=u_{1}(t)y_{1}(t)+u_{2}(t)y_{2}(t)$$
and\\

\begin{eqnarray}
&&u_{1}(t)=-\int \dfrac{y_{2}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
&&\quad\quad\ =-\int \dfrac{t\cdot (-2(t-1)e^{-t})}{(1-t)e^{t}}dt\\
&&\quad\quad\ =-2\int te^{-2t}dt\\
&&\quad\quad\ =(t+\dfrac{1}{2})e^{-2t}
\end{eqnarray}
\begin{eqnarray}
&&u_{2}(t)=\int \dfrac{y_{1}(t)g(t)}{W[y_{1},y_{2}](t)}dt\\
&&\quad\quad\ =\int \dfrac{e^{t}\cdot (-2(t-1)e^{-t})}{(1-t)e^{t}}dt\\
&&\quad\quad\ =2\int e^{-t}dt \\
&&\quad\quad\ =-2e^{-t}
\end{eqnarray}
Therefore,\\
$$Y(t)=(t+\dfrac{1}{2})e^{-2t}\cdot e^{t}+(-2e^{-t})\cdot t=(\dfrac{1}{2}-t)e^{-t}$$
Hence, the general solution:\\
\begin{eqnarray}
&&y(t)=y_{c}(t)+Y(t)\\
&&\quad\quad=c_{1}e^{t}+c_{2}t+(\dfrac{1}{2}-t)e^{-t}
\end{eqnarray}
Therefore, the particular solution of the given nonhomogeneous equation is\\
$$Y(t)=(\dfrac{1}{2}-t)e^{-t}$$
\end{document}

\textbf{Q3. $y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$,$y(0)=0$,$y^{\prime}(0)=0$}

\textbf{Ans:} let $y=e^{r x}$

$y^{\prime}=r e^{r x}$

$y^{\prime \prime}=r^{2} e^{r x}$

$r^{2} e^{r x}-5 r e^{r x}+6 e^{r x}=0$

$\left(r^{2}-5 r+6\right)=0$

$(r-2)(r-3)=0$

$r_{1}=2$

$r_{2}=3$

$y=a e^{r x}+c_{2} e^{r_{2} x}=u e^{2 x}+c_{2} e^{3 x}$

let $y=A \cos (2 x)+B \sin (2 x)$

$y^{\prime}=-2 A \sin (2 x)+2 B \cos (2 x)$

$y^{\prime \prime}=-4 A \cos (2 x)-4 B \sin (2 x)$

$y^{\prime \prime}-5 y^{\prime}+6 y=52 \cos (2 x)$

$-4 A \cos (2 x)-4 B \sin (2 x)-5(-2 A \sin (2 x)+2 B \cos (2 x))+6(A \cos (2 x)+B \sin (2 x))=52(\sin (2 x)$

$(-4 A-10 B+6 A) \cos (2 x)+(-4 B+10 A+6 B) \sin (2 x)=52 \cos (2 x)$

$(-10 B+2 A) \cos (2 x)+(10 A+2 B) \sin (2 x)=52 \cos (2 x)$

$\left\{\begin{array}{c}{-10 B+2 A=52} \\ {10 A+2 B=0}\end{array}\right.$ $\Rightarrow 5 A+B=0$

$\Rightarrow \quad B=-5 A$  Sub $B=-5 A$ into $-10 B+2 A=52$

$50 A+2 A=52$

$A=1$

$B=-5$

$\therefore y=\cos (2 x)-5 \sin (2 x)$

$\therefore y=c_{1} e^{2 x}+c_{2} e^{3 x}+\cos (2 x)-5 \sin (2 x)$

then $y^{\prime}=2 C_{1} e^{2 x}+3 C_{2} e^{3 x}-2 \sin (2 x)-10 \cos (2 x)$

$y(0)=0$  sub $x=0, \quad y=0$

$0=C_{1} e^{0}+\left(2 e^{0}+\cos (0)-5 \sin (0)\right)=C_{1}+C_{2}+1=0$

$C_{2}=-1-C_{1}$

$y^{\prime}(0)=0$  sub $x=0, y^{\prime}=0$
$0=2C_{1}  e^{0}+3\left(2 e^{0}-2 \sin (0)-10 \cos (0)=2 C_{1}+3 C_{2}-10=0\right.$

$2C_{1}+3 C_{2}=10$

Sub $c_{2}=-1-C_{1}$ info $2C_{1}+3 C_{2}=10$

$2C_{1}+\left(-3-3 C_{1}\right)=10$

$2 C_{1}-3-3 C_{1}=10$

$C_{1}=-13$

$C_{2}=-1+13=12$

$\therefore y=-13 e^{2 x}+12 e^{3 x}+\cos (2 x)-5 \sin (2 x)$

\textbf{\textbf{Ans:}}

$M=-y^{2} \sin (x y)$

$N=-x y \sin (x y)+2 \cos (x y)+3 y$

$M_{y}=-2 y \sin (x y)-y^{2} \cos (x y) \cdot x=-2 y \sin (x y)-x y^{2} \cos (x y)$

$N_{x}=-y \sin (x y)-x y^{2} \cos (x y)+2(-\sin (x y)) y+0=-y \sin (x y)-x y^{2}(\cos (x y)-2 y \sin (x y)$

$R_1=\frac{M_{y}-N_{x}}{M}=\frac{-2 y \sin (x y)-x y^{2} \cos (x y)+y \sin (x y)+x y^{2} \cos (x y)+2 y \sin (x y)}{-y^{2} \sin (x y)}=\frac{y \sin (x y)}{-y^{2} \sin (x y)}=-\frac{1}{y}$

$\mu=e^{-\int_{R_1} d y}=e^{\int \frac{1}{y} d y}=e^{\ln |y|}=y$

both muliply by $\mu$ $-y^{3} \sin (x y)+\left(-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}\right) y^{\prime}=0$

$\varphi(x, y)$

$\varphi_{x}=M=-y^{3} \sin (x y)$

$\varphi_{y}=N=-x y^{2} \sin (x y)+2 y \cos (x y)+3 y^{2}$

$\varphi=\int M d x=\int-y^{3} \sin (x y) d x=y^{2} \cos (x y)+h(y)$

$\varphi_{y}=2 y \cos (x y)+y^{2} x \sin (x y)+h^{\prime}(y)=N$

$h^{\prime}(y)=3 y^{2}$

$\int h^{\prime}(y) d y=\int 3 y^{2} d y=y^{3}$

$\varphi=y^{2} \cos (x y)+y^{3}=C$

put $y\left(\frac{\pi}{3}\right)=1$

$x=\frac{\pi}{3}$

$y=1$  into $ \varnothing $

$1^{2} \cos \left(\frac{\pi}{3} \times 1\right)+1^{3}=C=\frac{1}{2}+1=\frac{3}{2}$

$\therefore y^{2} \cos (x y)+y^{3}=\frac{3}{2}$

Here is my question and solution for the question I met in Quiz 3.