# Mostly Differential Equations and 3 others

• Apr 21st 2006, 09:45 AM
Mostly Differential Equations and 3 others
Hi, I am having trouble with differential equations . I studied these a long time ago and I'm having difficulty remembering them now. Also there are a coupla other problems that are not differential equations but I didnt want to open a seperate thread for them so I am including them here. I hope that's not a problem. Thank you so much

http://img53.imageshack.us/img53/961/a1q24il.jpg

http://img471.imageshack.us/img471/4061/a1q105st.jpg

http://img530.imageshack.us/img530/4381/a1q96sp.jpg

http://img374.imageshack.us/img374/5661/a1q133de.jpg
• Apr 21st 2006, 12:33 PM
topsquark
Umm...BIG problems! :D

Anyway:

$\displaystyle \frac{dy}{dx}-y=3e^x$

First solve the homogeneous equation:

$\displaystyle \frac{dy_h}{dx}-y_h=0$

We have that $\displaystyle m-1=0$ gives $\displaystyle m=1$ so

$\displaystyle y_h=Ae^x$

Now we need a particular solution. Try
$\displaystyle y_p=Bxe^{Cx}$
(Normally we'd perhaps try $\displaystyle y_p=Be^{Cx}$ first, but a quick check will show this doesn't work.)
So
$\displaystyle \frac{dy_p}{dx}=Be^{Cx}+BCxe^{Cx}$

Thus $\displaystyle \frac{dy_p}{dx}-y_p=Be^{Cx}+BCxe^{Cx}-Bxe^{Cx}=3e^x$

Thus $\displaystyle BC-B=0$ leaving $\displaystyle B=3$ and $\displaystyle C=1$, which checks in the initial condition.

Thus: $\displaystyle y_p=3xe^x$.

So $\displaystyle y=y_h+y_p = Ae^x+3xe^x$.

As you haven't listed any initial conditions, the problem is done.

-Dan
• Apr 21st 2006, 12:55 PM
ticbol
Quote:

Hi, I am having trouble with differential equations . I studied these a long time ago and I'm having difficulty remembering them now. Also there are a coupla other problems that are not differential equations but I didnt want to open a seperate thread for them so I am including them here. I hope that's not a problem. Thank you so much

Too many.

Why not post them separately. One per posting if possible.

I'd do the first and last problems here.

1.)Solve the Initial Value Problem.
(x^2 +1)(dy/dx) +y^2 +1 = 0; y(0) = 1.

(x^2 +1)(dy/dx) = -y^2 -1
(x^2 +1)dy = -(y^2 +1)dx
(x^2 +1)/dx = -(y^2 +1)/dy
Take the reciprocals of both sides,
dx/(x^2 +1) = -dy/(y^2 +1)
Integrate bot sides,
arctan(x) +C1 = -arctan(y) +C2
arctan(y) = -arctan(x) +C
Get the tan of both sides,
y = tan[C -arctan(x)] -----------(i)

Plug y(0) = 1 into (i),
Meaning, y=1 when x=0,
1 = tan[C -arctan(0)]
1 = (tanC -0)/(1 +(tanC)(0))
1 = tanC /1
tanC = 1
C = arctan(1) = pi/4 ----***

Substitute that into (i),
y = tan[pi/4 -arctan(x)]
y = [tan(pi/4) -tan(arctan(x))] / [1 +tan(pi/4)*tan(arctan(x))]
y = (1-x)/(1 +1*x)

--------------------
Find the name of the family of curves if the slope of the tangent line at any point (x,y) is (1-2x)/(1+y).

so, dy/dx = (1-2x)/(1+y).

Cross multiply,
(1+y)dy = (1-2x)dx
Integrate both sides,
y +(y^2)/2 +C1 = x -2(x^2)/2 +C2
Multiply both sides by 2, and combine 2C1 and 2C2 into C only,
2y +y^2 = 2x -2x^2 +C
2x^2 -2x +y^2 +2y -C = 0
(2x^2 -2x) +(y^2 +2y) -C = 0
1
• Apr 21st 2006, 12:58 PM
topsquark
$\displaystyle \frac{dy}{dx}-\frac{1}{x}y=-xy^2$.

In general a Bernoulli equation is of the form:

$\displaystyle \frac{dy}{dx}+P(x)y=Q(x)y^n$

In all cases the substitution $\displaystyle v=y^{1-n}$ is useful.

So, let $\displaystyle v=y^{1-2}=\frac{1}{y}$

Thus $\displaystyle y=\frac{1}{v}$ and $\displaystyle \frac{dy}{dx}=-\frac{1}{v^2} \frac{dv}{dx}$ so the differential equation becomes:

$\displaystyle -\frac{1}{v^2} \frac{dv}{dx}-\frac{1}{x}\frac{1}{v}=-x \left ( \frac{1}{v} \right ) ^2$

Or
$\displaystyle x \frac{dv}{dx}+v=x^2$

There are various and sundry methods to solve this equation. As it's a Cauchy-Euler equation, my favorite method is the substitution $\displaystyle x=e^t$, even though it's a bit of overkill for a first order equation.

$\displaystyle x=e^t$
$\displaystyle dx=e^t dt$
$\displaystyle \frac{d}{dx}=e^{-t}\frac{d}{dt}$

So the equation becomes:
$\displaystyle e^te^{-t}\frac{dv}{dt}+v=e^{2t}$

or
$\displaystyle \frac{dv}{dt}+v=e^{2t}$

The homogeneous solution of this equation is:
$\displaystyle m+1=0$ implies $\displaystyle m=-1$ implies
$\displaystyle v_h=Ae^{-t}$

And the particular solution:
$\displaystyle v_p=Be^{Ct}$
$\displaystyle \frac{dv_p}{dt}=BCe^{Ct}$
$\displaystyle BCe^{Ct}+Be^{Ct}=e^{2t}$

Gives $\displaystyle C=2$ and $\displaystyle BC+B=1$ or $\displaystyle B=1/3$.

So $\displaystyle v(t)=Ae^{-t}+\frac{1}{3}e^{2t}$

Converting t back to x:
$\displaystyle v(x)=\frac{A}{x}+\frac{1}{3}x^2$

Converting v back to y:
$\displaystyle y(x)=\frac{1}{\frac{A}{x}+\frac{1}{3}x^2}$

or
$\displaystyle y(x)=\frac{3x}{x^3+3A}$

-Dan
• Apr 21st 2006, 06:18 PM
Quote:

Originally Posted by ticbol
Why not post them separately. One per posting if possible.

I thought about doing that first. But then I felt that it would take up a little too much server space. If you like, I can delete the questions that havent been answered and post them seperately.

A Big thanks to topsquark and ticbol :D
• Apr 21st 2006, 06:49 PM
ticbol
Quote: