\magnification = 1200
\documentstyle{amsppt}
\baselineskip = 18 pt


{\bf 1.} Find the solution of the following initial value problem:
$$y'+\cos x \  y= \cos x,\ \  y(0)=2$$
\bigskip
\noindent
{\bf Solution} \newline
The integrating factor is $\mu(x) = e^ {\smallint cosxdx} = e^{sinx}$
\newline
After multiplying by $e^{sinx}$ the $eqn$ becomes $e^{sinx} y )^1 = cosx
e^{sinx}$ \newline 
Thus \newline
$e^{sinx} = \smallint cosxe^{sinx} dx + c = \smallint e^{sinx} d(sinx) + c   
 = e^{sinx} + c$  
\newline 
and $y(x) = 1 + ce^{-sinx}$ \newline
since $2 = y(0) = 1 + ce ^\circ = 1 + c$
\newline
We have $c=1$ and $ \boxed{y(x) =1 + e^{-sinx}}$
\bigskip
\noindent
{\bf 2.} A certain population increases at a rate proportional to the square 
root of the population present.  At the beginning the population was  100. 
One hour later it is 900.
\roster
	\item Find the population  $p(t)$  at any time.
	\item  Find the population four hours later;i.e.  find $p(4)$.
 \endroster


\bigskip
\noindent
{\bf Solution} \newline
$$\frac{dp}{dt} = r \sqrt p$$
$$\Rightarrow p^{{-\frac{1}{2}}}  dp = rdt \Rightarrow
{\frac{p^{-1/2+1}}{-1/2+ 1} = rt + c \Rightarrow \sqrt p
=  \frac{1}{2}(rt +c)$$

$$10 = \sqrt 100 = \sqrt {p(0)} = \frac1{2}(r(o) + c) =\Rightarrow c = 20$$ 
$$30 = \sqrt{900} = \sqrt {p(1)} =  {\frac1{2}}(r(1) + 20) \Rightarrow r = 40 \Rightarrow
P(t) = 100(2t + 1)^2$$ 
$$P(4) = 100(2(4 )+ 1)^2 = 100 (81)$$ 
 $$P(t) =100(2t +1)^2$$  
$$P(4) = 8,100$$

\newpage 
\bigskip
\noindent
{\bf 3.} A body with weigt  160 pounds falls from rest in a medium offering resistance
equal to $10v^2$. Find the speed  $v(t)$  of the body
 at any time $t$.  Also find $\lim_{t\to \infty}\,v(t).$ \newline
 

\bigskip
\noindent
{\bf Solution} \newline
$m {\frac {du}{dt}} = mg - 10u^2$, $m = {\frac w{g}} = {\frac{160}{32}} = 5$
\newline $ {\frac {du}{dt}} = 32 - 2u^2$ \newline
${\frac{du}{16 -v^2}} = 2dt \Rightarrow \int \frac{du}{16-v^2} = 2t + c$
\newline ${\frac1{8}} \int [{\frac1{4-u}} + {\frac1{4 + u}]} du = 2t + c$
\newline ${\frac1{8}} en(4 + u) - {\frac1{8}} en(4- u) = 2t + c$ \newline
$en{\frac{4 + u}{4- u}} = 16t + c$ \newline
At $t = 0$: $ln\frac{4 + 0}{4 - 0} = c$ or $ln1 = c = 0$ \newline
Thus ${\frac{4 + u}{4 - u}} = e^{16t}$ \newline
$u(t) = 4 {{\frac{1- e^{-16t}}{1- e^{-16t}}}}$

			

\bigskip
\noindent
{\bf 4.}  Find the general solution of the D.E.
$$y''-6y'+9y=e^{3x}+\sin x.$$
\bigskip
\noindent
{\bf Solution} First we solve the homogeneous equation
$$y'' - 6y' + 9y = 0.$$
To solve this we need to solve the quadratic $r^2 - 6r + 9 = 0.$  We find
$r = 3$ and the solutions to the homogeneous equation are 
$$y_{c} = c_{1}e^{3x} + c_{2} x e^{3x}$$ 
A particular solution of the non-honogeneous equation is given by
$$y_{p} = Ax^{2}e^{3x} + B sin X + C cos X$$
From the D.E. we obtain
$$A[2e^{3x} + xe^{3x} + 9x^{2}e^{3x}] - 6A [2xe^{3x} + 3x^{2}e^{3x}]
+9Ax^{2}e^{3x} + [-B sin X - C cos X] - 6[B cox X - C sin X] + 9[B sin X + C
Cos X] = e^{3x} + sin X$$
We must have
$2A=1, \ 8B + 6C = 1,  and \ 8C-6B = 0$

\bigskip
or $A = \frac {1}{2}, \ B = \frac {2}{25}, \ C= {3}{50}.$  Therefore, 
\bigskip
$$\boxed{{y = y_{c} + y_{p} = e^{3x}(c_{1} + c_{2}x + \frac {1}{2}x^{2})  +
\frac {2}{25} sin X + \frac {3}{50} Cos X}}$$

\bigskip

\noindent
{\bf 5.}  Solve the differential equation
$$y''-2y'+y=2e^x\ln x,\quad x>0.$$
\bigskip
\noindent
{\bf Solution}  The solution of the D.E. is given by $y=y_{c} + y_{p}$ where
$y_{c}$ is the solution to $y'' - 2y' + y = 0$ and $y_{p}$ is a particular
solution.  We have $r^{2} - 2r + 1 = (r-1)^{2} = 0$ or $r = 1.$  Thus, 
$$y_{c} = c_{1}e^{x} + c_{2}xe^{x}.$$

$y_{p}$ is given by $y_{p} = e^{x}u_{1} + xe^{x}u_{2}$ where

$$e^{x}u_{1}' + xe^{x}u_{2}' = 0$$
\bigskip
$$e^{x}u_{1}' + (x +1)e^{x}u_{2}' = 2e^{x}ln x$$

we have

$$u_1' = \frac{\vmatrix 0 & xe^x\\ 
2e^xlnx & (x+1)e^x\endvmatrix}{\vmatrix e^x &
xe^x\\  e^x & (x + 1)e^x \endvmatrix} = \frac{-2xe^{2x}ln
\ x}{e^2x} =-2x \  ln\ x$$

or, $$u_{1} = -\int2x \ ln \ x \ dx = - \int ln \ x \ d(x^2) = -x^2 \ ln \  x +
\int x^2 \frac {1}{x} dx$$

or

$$u_1 = -x^2 \ ln \ x + \frac{x^2}{2}$$

$$u_2' = \frac{\vmatrix e^x & 0\\
e^x & 2e^x\ lnx\endvmatrix }{\vmatrix e^x & xe^x\\
e^x & (x-11)e^x \endvmatrix} = \frac{2e^{2x}lnx}{e^{2x}} = 2\,ln \
x$$

or $$u_2 = 2 \in ln \, x \, dx = 2x \, ln \, x - 2 \in \frac{1}{x} x \, dx = 2x
\, ln \, x - 2x.$$

Therefore, $$y_p = e^x(-x^2 \,ln \, x + \frac{x^2}{2}) + xe^ax(2x \, ln \,x -
2x)$$

and
$$y=e^x[c_1 + c_2 x - x^2 ln\, x + \frac{x^2}{2} + 2x^2 \, ln \, x - 2x^2] $$

or
$$y=e^x[c_1 + c_2x + x^2 \, ln \, x - \frac {3}{2} x^2]$$

\bigskip
\noindent
{\bf 6.} Find the general solution to the D.E.
$$y'' - 4y' + 4y = 2e^{2x}$$ 
\bigskip
\noindent
{\bf Solution} The general solution of the D.E. is given by $y=y_c + y_p$
where $y_c$ is the general solution of the homogeneous equation and $y_p$
is a particular solution.  Since $r^2 - 4r - 5 = 0$ implies $r=2$ is a double
root, we have $$y_c = ce^{2x} + c_2 xe^2x$$
$y_p$ is of the form $y_p = Ax^2e^{2x}$.  From the D.E. we obtain
$$(2e^{2x} + 8xe^{2x} + 4x^2e^{2x}) - 4(2xe^{2x} + 2x^2e^{2x) +4x^2e^{2x} = 2e^{2x}$$

Therefore, $A=1$ and
$$ \boxed{y=(x^2 + c_2x + c_1)e^{2x}}$$

b)  Show that any solution to the D.E. is positive for all {\bf x} with $|x|$
large enough.
\bigskip
{\bf Solution}  By part (a), a solution to the D.E. is given by 
$$y(x) = (x^2 + bx + c) e^{2x}, \, b, c \in \Bbb R.$$

Since $e^{2x} > 0$ for all x and since for $x \neq 0$ we have
$$x^2 + bx + c = x^2 \left(1 + \frac{b}{x} + {c}{x^2}\right) \geq x^2 \left(1 -
\frac {|b|}{|x|} - \frac {|c|}{x^2}\right) \geq x^2 \left(1 - \frac{M}{|x|} - \frac
{M^2}{x^2}\right)$$ where $M= max \{|b|, \, |c|, \, 1\}.$


Thus, if $\frac{M}{|x|} < \frac{1}{2}$ or $2M < |x|$, then $x^2 + bx + c > x^2
\left(1 - \frac{1}{2} - \frac{1}{4} \right) = \frac{1}{4} x^2 > 0$. 

Thus, if $|x| > 2M$ then $y(x) > 0.$


\bigskip
\noindent
{\bf 7.} \underbar{Example} \,\, (Variation of Parameters) 1991 \newline
Find the general solution to the D.E.
$$y'' + y = \frac{1}{cos^3x}  \, , \, -\frac{n}{2}<x< \frac{n}{2}$$

\bigskip
\noindent
{\bf Solution} \, \, The general solution to D.E. is given by $y = y_c + y_p$
where $y_c$ is the general solution to the homogeneous solution, i.e.,
$y_c = c_1 Sin \, x + c_2 Cos \, x,$ and $y_p$ is a particular solution to
D.E., this solution is of the form $$y_p = u_1 Sin \, x + u_2 \, Cos \,x$$
where $u_1'$ and  $u_2'$ satisfy the system
$$Sin \, x \, u_1' + Cox \, x \, u_2' = 0$$
$$Cos \, x \, u_1' - Sin \, x \, u_2' = \frac{1}{cos^3x}$$

Thus  
$$u_1' = \frac{\vmatrix 0 & cos x\\
\frac{1}{cos^3x} & - sin x \endvmatrix }{\vmatrix sin
x & cos x  \\ cos x & -sin x \endvmatrix } =
\frac{1}{cos^2x}$$

and $$u_2 =\frac{\vmatrix sin x & 0\\
cos x & \frac{1}{cos^3x} \endvmatrix}{\vmatrix sin
x & cos x  \\ cos x & -sin x \endvmatrix} =
-\frac{sin \, x}{cos^3x}$$

Thus $$u_1 = \int \frac{dx}{cos^2x} = tan \, x$$

and $$u_2 = - \int \frac{sin x}{cos^3x}dx = \int \frac{d(cos x)}{cos^3x} =
\frac{(cos x)^{-3 +1}}{-3 + 1} = -\frac{1}{2 cos^2x}$$

The general solution to D.E. is 
$$\boxed{ y= c_1 sin x + c_2 cos x + tan x \, sin x - \frac {1}{2 cos^2 x}
cos \,x}$$ 

\bigskip
\noindent
{\bf 8.} 
\bigskip
\noindent
{\bf Solution.}  Example (similar).
Find the general solution to D.E.
$$y'' + 4y = \frac{1}{sin^32x}, -n < \, x \, < 0.$$
   
\bigskip
\noindent
{\bf 9.} 
Find the solution of the following initial value problem:
$$y'+4x^3y=x^3,\ \  y(0)=1$$
\bigskip
\noindent
{\bf Solution.} \, Integrating factor $$\mu (x) = e^{\int 4x^3dx} = e ^{x^4}$$
Thus, $\left(e^{x^4}y\right)' = x^3e^{x^4}$

$$\Rightarrow e^{x^4} = \int x^3e^{x^4} + c$$


$$\Rightarrow e^{x^4}y = \frac{1}{4}e^{x^4} + c$$

$$\Rightarrow y(x) = \frac{1}{4} + ce^{-x^4}$$

$$1=y(0)= \frac{1}{4} + c \, \, \Rightarrow  c=\frac{3}{4}$$

Therefore, $$y(x) = \frac{1}{4} + \frac{3}{4}e^{-x^4}$$

\bigskip
\noindent
{\bf 10.}  A certain population increases at a rate proportional to the cubic root of 
the population present.  In the $t=0$ the population is  1000. At $t=3$ later it is 27,000.
First  find a formula that gives the population  p(t)  at any time and then compute the 
 population at $t=45/8$.
\bigskip
\noindent
{\bf Solution.}  $$\frac {dP}{dt} = rp^\frac{1}{3}  \Rightarrow
p^{-\frac{1}{3}}dP = rdt \Rightarrow \frac{p^-\frac{1}{3} +
1}{-\frac{1}{3} + 1} = rt + c$$
\bigskip
$$\boxed{p^\frac{2}{3} = \frac{2}{3}(rt + c)} \cdot 1000^\frac{2}{3} =
\frac{2}{3}c \Rightarrow 100 = \frac{2}{3}c$$

$$\Rightarrow c = 150 \, \Rightarrow \boxed{P^\frac{2}{3}=\frac{2}{3}(rt
+ 150)}$$

$$27,000^\frac{2}{3} = \frac{2}{3}r \cdot 3 + 100 \Rightarrow 900 = 2r +
100 \Rightarrow \, r = 400$$

$$ \Rightarrow \boxed{ P^\frac{2}{3} = \frac{2}{3}(400t + 150)}$$

$$P(45/8) = \frac{2}{3}(400 \cdot \frac{45}{8} + 150) = 1,600$$ 


$$P(t) = \left[\frac{2}{3}(400t + 150)\right]^\frac{3}{2}$$

$$P(4\frac{5}{8}) = 1600^\frac{3}{2} = 40^{3} = 64,000$$

\bigskip
\noindent
{\bf 11.}

A man with a parachute jumps from a great hight. The combined weigt of man and parachute
is 160 pounds and the force of the air resistance is equal to $\frac{1}{2}v$ when the
parachute is closed. After 10 seconds he opens his parachute. Find his speed  $v(t)$ 
 at any time before he opens his parachute and at the time he opens his parachute. 
It is given that $1-e^{-1}\approxeq  0.632$. 

\bigskip
\noindent
{\bf Solution.}

$$m\frac{dv}{dt} = mg - \frac{1}{2}v$$

$$ \Rightarrow \frac{dv}{dt} + \frac{1}{2m}v = g$$

$$m = \frac{w}{g} = \frac {160}{32} = 5$$

$$\frac{dv}{dt} + 0.1v = 32$$

$$\Rightarrow \left(ve^{0.1t} \right)^{1} = 32e^{0.1t}$$

$$\Rightarrow  ve^{0.1t} = \frac{32}{0.1}e^{0.1t} + c$$

$$ \Rightarrow V(t) = 320 + ce^{-0.1t}, \, \, 0 = v(0) = 320 + c$$

$$V(t) = 320 \left( 1-e^{-0.1t} \right)$$

$$V(10) = (320) x (0.632) = 202.24 \, ft./sec.$$

$$ 
\boxed{\aligned m \,  \bullet \,  t &= 0\\
R &= - \frac{1}{2}v\\
m \, \bullet \,  t &=\\
w &= mg\\
\bullet \, t &= 10\\
\bullet x &= \\
\endaligned}$$


\bigskip
\noindent
{\bf 12.}  Find the general solution of the D.E.
$$y''-10y'+25y=0$$
Then find the solution with $y(0)=1$ and $y'(0)=9$.

\bigskip
\noindent
{\bf Solution.}
$$r^2 - 10r + 25 = (r - s)^2 = 0  \Rightarrow r = 5$$

$$y = c_1e^{5x} + c_2xe^{5x}$$

$$1 = y(0) = c_1$$

$$y' = 5c_1e^{5x} + c_2 \left(5xe^{5x} + e^{5x} \right)$$

$$y = y(0) = 5c_1 + c_2  \Rightarrow c_2 = 4$$

So,
$$y=c_1e^{5x} + c_2xe^{5x}$$

$$y=e^{5x} + 4xe^{5x}$$


\bigskip
\noindent
\bigskip
\noindent
{\bf 13.} Find a particular solution to the D.E:
$$y'' + y = \frac{1}{cos x}$$
Then find the general solution to the above D.E.

\bigskip
\noindent
{\bf Solution} \newline
We have that $y_1 = cos x$ and $y_2 = sin x$ are solutions to $y'' + y = 0.$ Then,
$y_p = y_1u_1 + y_2u_2$ where
$$
 \cases y_1u_1' + y_2u_2' =0\\
y_1'u_1' + y_2'u_2' = \frac{1}{cos x} \endcases  \, or \, \,  \cases cos  x u_1' + sin x u_2' = 0\\
-sin x u_1' + cos x u_2' = \frac {1}{cos x} \endcases$$

$$
u_1' = \frac{\left|\smallmatrix 0 & sin x\\ 
\frac{1}{coxx} & cox x\endsmallmatrix \right| }{\left|\smallmatrix cos x & sin x\\ -sin x & cos x
 \endsmallmatrix \right|} = \frac{-sin x}{cos x}  \Rightarrow u_1 = - \int \frac{sin x}{cos x}dx
= ln |cos x|$$

$$
u_2' = \left| \smallmatrix cos x & 0\\
-sin x & \frac{1}{cos x} \endsmallmatrix \right| = \, 1 \Rightarrow u_2 = x$$

So, $$y_p = cos \, ln|cos \,x| + x \, sin \, x$$
$$y = c_1 cos \, x + c_2 \, sin \, x + y_p$$



\bigskip
\noindent
{\bf 13.} Find a particular solution to the D.E:
$$y'' + y = \frac{1}{cos x}$$
Then find the general solution to the above D.E.

\bigskip
\noindent
{\bf Solution} \newline
We have that $y_1 = cos x$ and $y_2 = sin x$ are solutions to $y'' + y = 0.$ Then,
$y_p = y_1u_1 + y_2u_2$ where
$$
 \cases y_1u_1' + y_2u_2' =0\\
y_1'u_1' + y_2'u_2' = \frac{1}{cos x} \endcases  \, or \, \,  \cases cos  x u_1' + sin x u_2' = 0\\
-sin x u_1' + cos x u_2' = \frac {1}{cos x} \endcases$$

$$
u_1' = \frac{\vmatrix 0 & sin x\\ 
\frac{1}{coxx} & cox x\endvmatrix }{\vmatrix cos x & sin x\\ -sin x & cos x
 \endvmatrix} = \frac{-sin x}{cos x}  \Rightarrow u_1 = - \int \frac{sin x}{cos x}dx
= ln |cos x|$$

$$
u_2' =\vmatrix cos x & 0\\
-sin x & \frac{1}{cos x} \endvmatrix  = \Rightarrow u_2 = x$$
So,
$$y_p = cos \, x \, ln|cos \,| + x \, sin x$$
$$y = c_1 \, cos \, x + c_2 \, sin \, x + y_p$$.

\bigskip
\noindent
{\bf 14.} Find the general solution to the differential equation
$$xy'' - \left( 2x + 1 \right)y' + \left( x + 1 \right)y = 0$$
given that $y_1(x) = e^x$ is a solution. 

\bigskip
{\bf Solution.}  We look for a second independent solution $y_2$ in the form
$$y_2 = y_1(x)u(x) = e^x u(x)$$

We must have
$$x \left(e^xu \right)'' -  \left(2x + 1\right) \left( e^xu \right)' + \left( x + 1 \right) e^xu = 0$$
  or
$$x \left(e^xu'' + 2e^xu' + e^xu \right) - \left(2x + 1 \right) \left(e^xu' + e^xu \right) + \left(x + 1
\right)e^xu = 0$$ 
or
$$xu'' - u' = 0$$

Let $u=u'.$  Then $xu' = u$ or $ \frac{du}{u} = \frac{dx}{x}$ or $ln \, u = ln \, x$ or $u = x$ or $u' =
x$ or $u' = \frac{1}{2}x^2.$

Thus, $y_2(x) = \frac{1}{2}x^2e^x$ and the general solution is given by 

$$y(x) = e^x \left(c_1 + c_2x^2 \right).$$

\bigskip
\noindent
{\bf 15.}  \left(Variation of parameters \right).
 Find the general solution to the differential equations:

$$i. \, \,  y'' - y = \frac {1}{e^x + 1}$$

$$ii. \, \,  y'' - 2y' + y = \frac{e^x \, ln \, x}{x^2}$$

$$iii. \, \, xy'' - \left(2x +1 \right)y' + \left(x + 1 \right)y = g(x),$$ given that $y_1 = e^x$ and
$y_2 = \frac{x^2}{2}e^x$ are two independent solutions.$$

\bye
 




 

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