Toronto Math Forum

$\text(a)\\$
$r^2+4=0\\$
$\qquad r=\pm2i\\$
$\therefore y(t)=c_1 \cos{2t}+c_2 \sin{2t}\\$
$W=\begin{vmatrix}
\cos2t & \sin2t \\
-2\sin2t & 2\cos2t \\
\end{vmatrix}=2\\$
$W_1=\begin{vmatrix}
0 & \sin2t \\
1 & 2\cos2t \\
\end{vmatrix}=-\sin2t\\$
$W_2=\begin{vmatrix}
\cos2t & 0 \\
-2\sin2t & 1 \\
\end{vmatrix}=\cos2t\\$
$y_c(t)=\cos2t \int \frac{-\sin2s·\frac{1}{\cos^2{s}}}{2}ds+\sin2t \int \frac{\cos2s·\frac{1}{cos^2{s}}}{2}ds\\$
$\qquad = \frac{1}{2}\cos2t \int \frac{-2sins·coss}{cos^2{s}}ds+\frac{1}{2}\sin2t\int\frac{2cos^2{s}-1}{cos^2{s}}ds\\$
$\qquad =-\cos2t \int tans ds+ \frac{1}{2}\sin2t \int (2-\sec^2{s})ds\\$
$\qquad = \cos2t·\ln (\cos t) + \frac{1}{2}(2t-\tan t)\\$
$y(t)=c_1 \cos{2t}+c_2 \sin{2t}+ \cos2t·\ln (\cos t) + \frac{1}{2}(2t-\tan t)\\$
$\text(b)\\$
$y^{\prime}(t)=-2c_1\sin2t =2c_2\cos2t-2\sin2t·\ln\cos(t)-\cos2t·tant+\cos2t(2t- tant)+\frac{1}{2}\sin2t(2-\sec^2{t})\\$
$y(0)=c_1+\ln1=0 \implies c_1=0\\$
$y^{\prime}(0)=2c_2+(0-\tan0)=0 \implies c_2=0\\$
$\therefore y(t)=\cos2t·\ln (\cos t)+\frac{1}{2}\sin2t(2t-tant)$

$\det(A-\lambda I)=0\\$
$\begin{vmatrix}
1-\lambda & 3 \\
-2 &  -3-\lambda\\
\end{vmatrix}=(1-\lambda)(-3-\lambda)+6=0\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\quad \therefore \lambda=-1\pm \sqrt 2 i\\$
$\text{when $\lambda =-1+ \sqrt 2 i$},\\$
$
\begin{pmatrix}
2-\sqrt 2 i & 3 \\
-2 & -2-\sqrt 2 i
\end{pmatrix}= \begin{pmatrix}
2+\sqrt 2 i \\
-2
\end{pmatrix}\\$
$e^{(-1+ \sqrt 2 i)t}\begin{pmatrix}
2+\sqrt 2 i \\
-2
\end{pmatrix}=e^{-t}\begin{pmatrix}
2+\sqrt 2 i \\
-2
\end{pmatrix}(\cos\sqrt2t+i\sin\sqrt 2t)\\$
$\qquad\qquad\qquad\qquad =e^{-t}\begin{pmatrix}
2\cos\sqrt 2t+2i\sin\sqrt2t+\sqrt2i\cos\sqrt2t-\sqrt2\sin\sqrt2t \\
-2\cos\sqrt2t-2i\sin\sqrt2t
\end{pmatrix}\\$
$\qquad\qquad\qquad\qquad =e^{-t}i\begin{pmatrix}
2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\
-2\sin\sqrt2t
\end{pmatrix}+e^{-t}\begin{pmatrix}
2\cos\sqrt 2t -\sqrt2\sin\sqrt2t\\
-2\cos\sqrt2t
\end{pmatrix}\\$
$\therefore x(t)=c_1e^{-t}\begin{pmatrix}
2\sin\sqrt2t+\sqrt2\cos\sqrt2t\\
-2\sin\sqrt2t
\end{pmatrix}+c_2e^{-t}\begin{pmatrix}
2\cos\sqrt 2t -\sqrt2\sin\sqrt2t \\
-2\cos\sqrt2t
\end{pmatrix}\\$

$\text{(a)}\\$
$M=y+3y^2e^{2x}\qquad M_{y}=\frac{\partial}{\partial y}M=1+6ye^{2x}\\$
$N=1+2ye^{2x}\quad\quad N_{x}=\frac{\partial}{\partial x}N=4ye^{2x}\\$
$\text{Since $M_{y}\neq N_{x}$, the given differential equation is not exact.}\\ $

$R_2=\frac{M_y-N_x}{N}=\frac{1+6ye^{2x}-4ye^{2x}}{1+2ye^{2x}}=\frac{1+2ye^{2x}}{1+2ye^{2x}}=1\\$
$\mu=e^{\int R_2dx}=e^{\int1dx}=e^x\\$
$(e^{x}y+3y^2e^{3x})+(e^x+2ye^{3x})y^{\prime}=0\\ \\$

$\text{$\exists \psi{(x,y)}$ such that $\psi_{x}=M$}\\$
$\qquad\quad\psi{(x,y)}=\int {(e^{x}y+3y^2e^{3x})dx}\\$
$\qquad\qquad\qquad =e^xy+y^2e^{3x}+h(y)\\$
$\qquad\quad\psi_{y}=e^x=2ye^{3x}+h^{\prime}(y)=N\\$
$\qquad\quad h^{\prime}(y)=0\\$
$\qquad\quad h(y)=C\\$
$\text{and we have}\\$
$\qquad\quad\psi{(x,y)}=e^xy+y^2e^{3x}=C\\$

$\text{(b)}\\$
$\text{Since y(0)=1}\\$
$e^0·1+1^2·e^{3·0}=C\\$
$C=2\\$
$\text{Thus,}\\$
$e^xy+y^2e^{3x}=2\\$

$\text{(a)}\\$
$r^2-2r-3=0\\$
$(r+1)(r-3)=0\\$
$r=-1, r=3\\$
$\text{Homogeneous Equation: $y_c(x)=c_1e^{-x}+c_2e^{3x}$}\\ \\$

$\text{Since we know $\cosh{x}=\frac{e^{x}+e^{-x}}{2} = \frac{1}{2}e^x+\frac{1}{2}e^{-x}$}\\$
$\text{Therefore,}\\$
$y^{\prime\prime}-2y^{\prime}-3y=8e^x+8e^{-x}\\$
$Y(x)=Ae^x+Bxe^{-x}\\$
$Y^{\prime}(x)=Ae^x+Be^{-x}-Bxe^{-x}\\$
$Y^{\prime\prime}(x)=Ae^x-Be^{-x}-Be^{-x}+Bxe^{-x}\\ $
$Ae^x-2Be^{-x}+Bxe^{-x}-2Ae^x-2Be^{-x}+2Bxe^{-x}-3Ae^x-3Bxe^{-x} = 8e^x+8e^{-x}\\$
\begin{cases}
  (A-2A-3A)e^{x}=8e^{x}\\\
  (-2B-2B)e^{-x}=8e^{-x}
   \end{cases}
\begin{cases}
  A=-2\\
 B=-2
   \end{cases}
$Y(x)=-2e^x-2xe^{-x}\\$
$\text{General Solution:}\\$
$y(t)=c_1e^{-x}+c_2e^{3x}-2e^x-2xe^{-x}\\$


$\text{(b)}\\$
$y^{\prime}=-c_1e^{-x}+3c_2e^{3x}-2e^{x}-2e^{-x}+2xe^{-x}\\$
$y(0)=0\Rightarrow c_1+c_2-2=0\\$
$y^{\prime}(0)=0\Rightarrow -c_1+3c_2-2-2=0\\$
\begin{cases}
  c_1=\frac{1}{2}\\
 c_2=\frac{3}{2}
   \end{cases}

$\text{Thus,}$
$y(t)=\frac{1}{2}e^{-x}+\frac{3}{2}e^{3x}-2e^x-2xe^{-x}$

$\text{(a)}\\$
$y^{\prime\prime}-\frac{2}{x}y^{\prime}+\frac{x^2+2}{x^2}y=0\\$
$W=ce^{-\int {p(x)dx}}=ce^{2\int {\frac{1}{x}dx}}=cx^2\\$
$\text{Let $c=1, W=x^2$ }\\ \\$

$\text{(b)}\\$
$\text{Verify that $y_1(x) = x\cos x$ is the solution}\\$
$y_1(x) = x\cos x\\$
$y_1^{\prime}(x) = \cos x-x\sin x\\$
$y_1^{\prime\prime}(x)= -\sin x-\sin x -x\cos x = -2\sin x-x\cos x\\$
$\text{Substitute these into the given equation: $x^2y^{\prime\prime}-2xy^{\prime}+(x^2+2)y=0$}\\$
$\quad x^{2}(-2\sin x-x\cos x)-2x(\cos x-x\sin x)+(x^{2}+2)(x\cos x)=0\\$
$-2x^{2}\sin x-x^{3}\cos x-2x\cos x+2x^{x}\sin x+x^{3}\cos x+2x\cos x=0\\$
$\text{Therefore, $y_1(x)=x\cos x$ is the solution.}\\ \\$
$W=\begin{vmatrix}
x\cos x & y_2 \\
\cos x-x\sin x & y_2^{\prime} \\
\end{vmatrix}=x^2\\$

$(x\cos x)y_2^{\prime}-(\cos x-x\sin x)y_2=x^2\\$
$y_2^{\prime}- \frac{\cos x-x\sin x}{x\cos x}y_2=\frac{x}{\cos x}\\$
$\mu = e^{\int {p(x)dx}}=e^{-\int {\frac{\cos x-x\sin x}{x\cos x}dx}} \qquad\text{By substitution, $u=x\cos x, du=(\cos x-x\sin x)dx$}\\$
$\qquad\qquad\quad =e^{-\int \frac{1}{u}}du\\$
$\qquad\qquad\quad=e^{-lnu}\\$
$\qquad\qquad\quad=\frac{1}{u}\\$
$\qquad\qquad\quad=\frac{1}{x\cos x}\\ \\$

$\frac{1}{x\cos x}y_2^{\prime}-{\frac{\cos x-x\sin x}{x^{2}\cos^{2} x} y_2}=\frac{1}{\cos^{2} x}\\$
$\frac{1}{x\cos x}y_2=\int \frac{1}{\cos^{2} x}dx\\$
$\qquad\quad=\int \sec^{2} xdx\\$
$\qquad\quad=\tan x+c\\$
$y_2=x\cos x·\tan{x}+x\cos x·c\\$
$\quad=x\cos x·\frac{\sin x}{\cos x}+x\cos x·c\\$
$\text{Let $c=0, y_2=x\sin x$}\\$
$\text{Therefore,}\\$
$y=c_1x\cos x+c_2x\sin x\\$

$\text{(c)}\\$
$y^{\prime}=c_1\cos x-c_1x\sin x+c_2\sin x+c_2x\cos x\\$
$y(\frac{\pi}{2})=1\Rightarrow c_1·(\frac{\pi}{2})·\cos (\frac{\pi}{2})+c_2·(\frac{\pi}{2})·\sin (\frac{\pi}{2})=1 \\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad c_2=\frac{2}{\pi}\\$
$y^{\prime}(\frac{\pi}{2})=0\Rightarrow -c_1·\frac{\pi}{2}·1+c_2·1=0\\$
$\quad\qquad\qquad\qquad\qquad\qquad c_1=\frac{4}{\pi^2}\\$
$\text{Thus,}\\$
$y=\frac{4}{\pi^2}x\cos x+\frac{2}{\pi}x\sin x$

No reason to post after this. V.I.

$\text{Find the Wronskian of the given pair of functions}$
$\text{$cos^2(x)$, $1+cos(2x)$}$

$$\quad
   W= \begin{vmatrix}
     cos^2(x) & 1+cos(2x) \\
     -2cos(x)sin(x) & -2sin2x \\
    \end{vmatrix}
$$
$$   
= \begin{vmatrix}
     cos^2(x) & 1+cos(2x) \\
     -sin2x & -2sin2x \\
    \end{vmatrix}
$$
$$\qquad\qquad\qquad\quad
=-2cos^2(x)sin(2x)-(-sin(2x))(1+cos(2x))
$$
$$\qquad\quad
=(-sin2x)(2cos^2(x)-1-cos(2x))
$$
$$\qquad\qquad\quad
=(-sin2x)(2cos^2(x)-1-2cos^2(x)+1)
$$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
=(-sin2x)·0
\\$
$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad
=0
$