# Thread: Ordinary differential equations

1. ## Ordinary differential equations

Ok, can someone check my qorking is sound for these please? ta

question is to find the solution of:

dy/dx - (y^2)/(x+1)=0

For which y=1 at x=0

i've got y^-2 dy = (x+1)^-1 dx

Then integrated both sides to get

-y^-1 = ln|x+1| + constant

which ends with the equation

y = -(ln|x+1|-1)^-1.

number 2

general solution of:

dx/dt + x =e^t

ive got it as:

Sdx + S(x)dt=S(e^t) dt

To give an answer of

x(t+1) = e^t +c

which doesnt seem right or is it?

Final one which i could do with a few pointers on is

(dw/dx)sin x -wcosx = xsin^2 x

find an integrating factor u(x)

2. Originally Posted by macabre
number 2

general solution of:

dx/dt + x =e^t

ive got it as:

Sdx + S(x)dt=S(e^t) dt

To give an answer of

x(t+1) = e^t +c

which doesnt seem right or is it?
x=(1/2)exp(t) is a particular integral of dx/dt+x=exp(t), as can be
seen by differentiating it and substituting it and its derivative into
the equation.

The general solution is a particular integral plus a solution to the
homogeneous equation dx/dt+x=0. Which is Aexp(-t).

So the general solution is: x(t)=Aexp(-t)+(1/2)exp(t).

RonL

3. ## could you clarify this please?

I'm afraid i dont understand your reasoning, would i tbe possible to provide a step by step working? also, what do you mean by Aexp? thanks a lot

4. Originally Posted by macabre
Ok, can someone check my qorking is sound for these please? ta

question is to find the solution of:

dy/dx - (y^2)/(x+1)=0

For which y=1 at x=0
Okay,

y'=y^2/(x+1)
Thus,
(1/y^2)y'=1/(x+1)
This is a seperable equation,
INTEGRAL (1/y^2)y' dx= INTEGRAL 1/(x+1) dx
Thus,
-1/y=ln|x+1|+C
Thus,
y=1/(C-ln|x+1|)
When x=0 we have,
1=1/(C-ln 1)
Thus,
1=1/C
Thus,
C=1
Thus,
y=1/(1-ln|x+1|)

5. Originally Posted by macabre

number 2

general solution of:

dx/dt + x =e^t
The equation, first order linear diophantine equation,
x'+x=0
Has a solution, we the entire number line.
Use the formula for solving 'first order linear',
x=1/u(x) INTEGRAL u(x) 0 dx
Thus,
x=C/u(x)
Where u(x) is integrating factor, which is,
u(x)=exp ( INTEGRAL 1 dx)=exp (x)=e^x
Thus,
x=Ce^(-t)
Now,
To solve the non-homogenous diophantine equation,
x'+x=e^t
We need a particular solution,
We search for solutions of the form,
x=Ae^t
Thus,
x'=Ae^t and x=Ae^t
Thus,
Ae^t+Ae^t=e^t
Thus,
2Ae^t=e^t
Thus,
2A=1
Thus,
A=1/2
Thus, all solutions to the equation,
x'+x=e^t are,
Ce^(-t)+1/2e^t

6. Originally Posted by ThePerfectHacker
The equation, first order linear diophantine equation,
x'+x=0

eherem.. differential

Has a solution, we the entire number line.
Use the formula for solving 'first order linear',
x=1/u(x) INTEGRAL u(x) 0 dx
Thus,
x=C/u(x)
Where u(x) is integrating factor, which is,
u(x)=exp ( INTEGRAL 1 dx)=exp (x)=e^x
Thus,
x=Ce^(-t)
At this stage you should know that a solution to:

dx/dt=-x

is of the form x(t)=A exp(-t)

Now,
To solve the non-homogenous diophantine equation,
x'+x=e^t
We need a particular solution,
We search for solutions of the form,
x=Ae^t
It is fairly clear that if we have a linear differential equation:

A(x)dx/dt+B(x)=G(t) .. (1)

That if x1 is some solution to this, and x2 is a solution to
the homogeneous equation:

A(x)dx/dt+B(x)=0 .. (2),

then x1+x2 is also a solution to (1). Hence the general solution to
(1) can be written in the form of the sum of a particular solution to
(1) amd the general solution to (2).

Thus,
x'=Ae^t and x=Ae^t
Thus,
Ae^t+Ae^t=e^t
Thus,
2Ae^t=e^t
Thus,
2A=1
Thus,
A=1/2
Thus, all solutions to the equation,
x'+x=e^t are,
Ce^(-t)+1/2e^t
RonL

7. Originally Posted by CaptainBlank
eherem.. differential
Too much number theory
---
Was that post too correct me? Or to explain me? It seems to explain me, because I think I did everyting correct.