Re: Problems...

• To: mathgroup at smc.vnet.net
• Subject: [mg73763] Re: Problems...
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Wed, 28 Feb 2007 04:24:49 -0500 (EST)
• References: <es131l\$nts\$1@smc.vnet.net>

```Hi.

Do you believe that this forum is the best place for solving your
homework?

of the
exercise or ask for better Mathematica code or even mathematical
details but
don't you feel what is you ask is too much?

There is a reason that your professor give you these exercises and it
is not good to cheat him.

For example here you could have already solved your problems by hand
(both of the exercises
are trivial). You could use Mathematica to verify your solutions, plot
them, investigate them etc.

You are bored and don't want work with pen and paper?
You  could use Mathematica and your BRAIN in order to solve these
equations.

After that you can consult to this forum asking for better code/
implementations/insight and many
many other things.

Actually there persons in this forum who can show you how you can use
Mathematica in order
even to cook with it.

But you should have prepared the "food" to be cooked by yourself!

Anyway...

There are many people in this forum that "could solve the below
problems".
Here are my attempts; far for regarding perfect, but here goes:

(1)

a) Mathematica's solution of dy/dx = y(1 - y)/(y + 1)

In[186]:=
Clear["Global`*"]
DSolve[deq1 = Derivative[1][y][x] == y[x]*((1 -
y[x])/(y[x] + 1)), y[x], x](*solution of DE*)
y[x_] = y[x] /. %[[2]](*y is always positive*)
Simplify[deq1](*check*)
Reduce[y[0] == 2, C[1], Reals] (*evaluation of arbitrary constant*)
ToRules[%]
ypart[x_] = y[x] /. %(*particular solution*)
Plot[Evaluate[%], {x, -3, 3}, Axes -> False, Frame -> {True, True,
False, False}, PlotStyle -> Blue];

Out[187]=
{{y[x] -> ((1/2)*(2*E^x + E^C[1] - E^(C[1]/2)*Sqrt[4*E^x + E^C[1]]))/
E^x},
{y[x] -> ((1/2)*(2*E^x + E^C[1] + E^(C[1]/2)*Sqrt[4*E^x + E^C[1]]))/
E^x}}

Out[188]=
((1/2)*(2*E^x + E^C[1] + E^(C[1]/2)*Sqrt[4*E^x + E^C[1]]))/E^x

Out[189]=
True

Out[190]=
C[1] == -Log[2]

Out[191]=
{C[1] -> -Log[2]}

Out[192]=
((1/2)*(1/2 + 2*E^x + Sqrt[1/2 + 4*E^x]/Sqrt[2]))/E^x

b) Integration method

In[27]:=
x == Integrate[(y + 1)/(y*(1 - y)), y] + Log[c]
% //. {(b_)*Log[a_] :> Log[a^b], Log[b_] + Log[c_] :> Log[b*c]}
(Exp[#1] & ) /@ %
Reduce[% && y > 0 && c > 0, y, Reals]
y[x_] = %[[3,2,2]]
Simplify[Derivative[1][y][x] == y[x]*((1 - y[x])/(y[x] + 1))]
Solve[y[0] == 2, c]
y[x] /. %[[1]]
Plot[Evaluate[%], {x, -3, 3}, Axes -> False, Frame -> {True, True,
False, False}, PlotStyle -> Blue];

Out[27]=
x == Log[c] - 2*Log[-1 + y] + Log[y]

Out[28]=
x == Log[(c*y)/(-1 + y)^2]

Out[29]=
E^x == (c*y)/(-1 + y)^2

Out[30]=
E^x > 0 && c > 0 && (y == ((1/2)*(c + 2*E^x))/E^x - (1/2)*Sqrt[(c^2 +
4*c*E^x)/E^(2*x)] ||
y == ((1/2)*(c + 2*E^x))/E^x + (1/2)*Sqrt[(c^2 + 4*c*E^x)/E^(2*x)])

Out[31]=
((1/2)*(c + 2*E^x))/E^x + (1/2)*Sqrt[(c^2 + 4*c*E^x)/E^(2*x)]

Out[32]=
True

Out[33]=
{{c -> 1/2}}

Out[34]=
(1/2)*Sqrt[(1/4 + 2*E^x)/E^(2*x)] + ((1/2)*(1/2 + 2*E^x))/E^x

2) Mathematica can solve with DSolve this equation.
For the substitution method set y/x=u and proceed with the integration
method.

Enough said!

Kind Regards
Dimitris

=CF/=C7 anooja m g =DD=E3=F1=E1=F8=E5:
> Hi,
>
>   It will be really helpful if someone could solve the below problems.
>
>
>   1) Given that y is always positive and y(0) = 2. By direct integratio=
n method, find the
> particular solution of the differential equation
>   dy/dx=y(1-y)/(y+1)
>   Hence show that y2-y(2+1/2e-x)+1=0
>
>   2)Using substitution method, find the general solution of the following=
homogeneous
> differentiation equation
>   dy/dx=(y/x)/1+(y/x)2
>   Hence find the particular solution if y(4)=2.
>
>
>   Regards,
>   Anooja

```

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